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Lemma 5.1.5 Given \( {z}_{0} \) in \( \mathbb{D} \) and \( {z}_{1} \) on \( {\mathbb{S}}_{\infty }^{1} \) there exists a transformation in \( \mathcal{H} \) that sends \( {z}_{0} \) to 0 and \( {z}_{1} \) to 1 .
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Therefore, one may view any transformation in \( \mathcal{H} \) as the composition of two inversions about clines orthogonal to \( {\mathbb{S}}_{\infty }^{1} \) . Moreover, these maps may be categorized according to whether the two clines of inversion intersect zero times, once, or twice. In Figure 5.1.6 we illustrate these three cases. In each case, we build a transformation \( T \) in \( \mathcal{H} \) by inverting about the solid clines in the figure (first about \( {L}_{1} \), then about \( {L}_{2} \) ). The figure also tracks the journey of a point \( z \) under these inversions, first to \( {z}^{\prime } \) by inverting about \( {L}_{1} \), then onto \( T\left( z\right) \) by inverting \( {z}^{\prime } \) about \( {L}_{2} \) . The dashed clines in the figure represent some of the clines of motion, the clines along which points are moved by the transformation. Notice these clines of motion are orthogonal to both clines of inversion.
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No
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Any transformation \( T \) in the group \( \mathcal{H} \) has the form \( T\left( z\right) = {e}^{i\theta }\frac{z - {z}_{0}}{1 - {\bar{z}}_{0}z} \) where \( \theta \) is some angle, and \( {z}_{0} \) is the point inside \( \mathbb{D} \) that gets sent to 0.
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Assume \( {z}_{0} \) is in \( \mathbb{D},{z}_{1} \) is on the unit circle \( {\mathbb{S}}_{\infty }^{1} \), and that the map \( T \) in \( \mathcal{H} \) sends \( {z}_{0} \mapsto 0,{z}_{0}^{ * } \mapsto \infty \) and \( {z}_{1} \mapsto 1 \) . Using the cross ratio, \( T\left( z\right) = \left( {z,{z}_{1};{z}_{0},{z}_{0}^{ * }}\right) = \frac{{z}_{1} - {z}_{0}^{ * }}{{z}_{1} - {z}_{0}} \cdot \frac{z - {z}_{0}}{z - {z}_{0}^{ * }}. \) But \( {z}_{0}^{ * } = \frac{1}{\overline{{z}_{0}}} \), so \( T\left( z\right) = \frac{{z}_{1} - 1/\overline{{z}_{0}}}{{z}_{1} - {z}_{0}} \cdot \frac{z - {z}_{0}}{z - 1/\overline{{z}_{0}}} = \frac{\overline{{z}_{0}}{z}_{1} - 1}{{z}_{1} - {z}_{0}} \cdot \frac{z - {z}_{0}}{\overline{{z}_{0}}z - 1}. \) Now, the quantity \( \frac{\overline{{z}_{0}}{z}_{1} - 1}{{z}_{1} - {z}_{0}} \) is a complex constant, and it has modulus equal to 1 . To see this, observe that since \( 1 = {\left| {z}_{1}\right| }^{2} = {z}_{1}\overline{{z}_{1}} \), \( \frac{\overline{{z}_{0}}{z}_{1} - 1}{{z}_{1} - {z}_{0}} = \frac{\overline{{z}_{0}}{z}_{1} - {z}_{1}\overline{{z}_{1}}}{{z}_{1} - {z}_{0}} = \frac{-{z}_{1}\left( \overline{{z}_{1} - {z}_{0}}\right) }{{z}_{1} - {z}_{0}} \) Since \( \left| {z}_{1}\right| = 1 \) and, in general \( \left| \beta \right| = \left| \bar{\beta }\right| \) we see that this expression has modulus 1, and can be expressed as \( {e}^{i\theta } \) for some \( \theta \) . Thus, if \( T \) is a transformation in \( \mathcal{H} \) it may be expressed as \( T\left( z\right) = {e}^{i\theta }\frac{z - {z}_{0}}{1 - {\bar{z}}_{0}z} \) where \( \theta \) is some angle, and \( {z}_{0} \) is the point inside \( \mathbb{D} \) that gets sent to 0.
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Yes
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Theorem 5.2.3 There exists a unique hyperbolic line through any two distinct points in the hyperbolic plane.
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Proof. Let \( p \) and \( q \) be arbitrary points in \( \mathbb{D} \) . Construct the point \( {p}^{ * } \) symmetric to \( p \) with respect to the unit circle, \( {\mathbb{S}}_{\infty }^{1} \) . Then there exists a cline through \( p, q \) and \( {p}^{ * } \), and this cline will be orthogonal to \( {\mathbb{S}}_{\infty }^{1} \), so it gives a hyperbolic line through \( p \) and \( q \) . Since there is just one cline through \( p, q \) and \( {p}^{ * } \), this hyperbolic line is unique.
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Yes
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Theorem 5.2.4 Any two hyperbolic lines are congruent in hyperbolic geometry.
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Proof. We first show that any given hyperbolic line \( L \) is congruent to the hyperbolic line on the real axis. Suppose \( p \) is a point on \( L \), and \( v \) is one of its ideal points. By Lemma 5.1.5 there is a transformation \( T \) in \( \mathcal{H} \) that maps \( p \) to 0, \( v \) to 1, and \( {p}^{ * } \) to \( \infty \) . Thus \( T\left( L\right) \) is the portion of the real axis inside \( \mathbb{D} \), and \( L \) is congruent to the hyperbolic line on the real axis. Since any hyperbolic line is congruent to the hyperbolic line on the real axis, the group nature of \( \mathcal{H} \) ensures that any two hyperbolic lines are congruent.
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Yes
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Theorem 5.2.6 Given a point \( {z}_{0} \) and a hyperbolic line \( L \) not through \( {z}_{0} \), there exist two distinct hyperbolic lines through \( {z}_{0} \) that are parallel to \( L \) .
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Proof. Consider the case where \( {z}_{0} \) is at the origin. The line \( L \) has two ideal points, call them \( u \) and \( v \), as in Figure 5.2.7. Moreover, since \( L \) does not go through the origin, Euclidean segment \( {uv} \) is not a diameter of the unit circle. Construct one Euclidean line through 0 and \( u \), and a second Euclidean line through 0 and \( v \) . (These lines will be distinct because \( {uv} \) is not a diameter of the unit circle.) Each of these lines is a hyperbolic line through 0 , and each shares exactly one ideal point with \( L \) . Thus, each is parallel to \( L \) . The fact that the result holds for general \( {z}_{0} \) is left as an exercise.
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No
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Theorem 5.2.11 Given any points \( p \) and \( q \) in \( \mathbb{D} \), there exists a hyperbolic circle centered at \( p \) through \( q \) .
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Proof. Given \( p, q \in \mathbb{D} \), construct \( {p}^{ * } \), the point symmetric to \( p \) with respect to \( {\mathbb{S}}_{\infty }^{1} \) . Then by Exercise 3.5.15 there exists a type II cline of \( p \) and \( {p}^{ * } \) that goes through \( q \) . This type II cline lives within \( \mathbb{D} \) because \( {\mathbb{S}}_{\infty }^{1} \) is also a type II cline of \( p \) and \( {p}^{ * } \), and distinct type II clines cannot intersect. This type II cline is the hyperbolic circle centered at \( p \) through \( q \) .
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No
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Example 5.3.4 The distance between two points.
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For instance, suppose \( p = \frac{1}{2}i, q = \frac{1}{2} + \frac{1}{2}i, z = {.95}{e}^{{i5\pi }/6} \) and \( w = - {.95} \) . Then \( {d}_{H}\left( {p, q}\right) \approx {1.49} \) units, while \( {d}_{H}\left( {z, w}\right) \approx {4.64} \) units.
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No
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Corollary 5.3.8 All points on a hyperbolic circle centered at \( p \) are equidistant from \( p \) .
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Proof. Suppose \( u \) and \( v \) are on the same hyperbolic circle centered at \( p \) . That is, these points are on the same type II cline with respect to \( p \) and \( {p}^{ * } \), so there exists a hyperbolic rotation that fixes \( p \) and maps \( u \) to \( v \) . Thus, \( {d}_{H}\left( {p, u}\right) = {d}_{H}\left( {T\left( p\right), T\left( u\right) }\right) = {d}_{H}\left( {p, v}\right) \) . It follows that any two points on the hyperbolic circle centered at \( p \) are equidistant from \( p \) .
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Yes
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Theorem 5.3.9 Hyperbolic lines are geodesics; that is, the shortest path between two points in \( \left( {\mathbb{D},\mathcal{H}}\right) \) is along the hyperbolic segment between them.
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Proof Sketch: We first argue that the geodesic from 0 to a point \( c \) on the positive real axis is the real axis itself.\n\nSuppose \( \mathbf{r}\left( t\right) = x\left( t\right) + {iy}\left( t\right) \) for \( a \leq t \leq b \), is an arbitrary smooth curve from 0 to \( c \) (so \( \mathbf{r}\left( a\right) = 0 \) and \( \mathbf{r}\left( b\right) = c \) ).\n\nSuppose further that \( x\left( t\right) \) is nondecreasing (if our path backtracks in the \( x \) direction, we claim the path cannot possibly be a geodesic). Then\n\n\[ \mathcal{L}\left( \mathbf{r}\right) = {\int }_{a}^{b}\frac{2}{1 - \left\lbrack {{\left( x\left( t\right) \right) }^{2} + {\left( y\left( t\right) \right) }^{2}}\right\rbrack }\sqrt{{\left( {x}^{\prime }\left( t\right) \right) }^{2} + {\left( {y}^{\prime }\left( t\right) \right) }^{2}}{dt}. \]\n\nThe hyperbolic line segment from 0 to \( c \) can be parameterized by \( {\mathbf{r}}_{0}\left( t\right) = \) \( x\left( t\right) + {0i} \) for \( a \leq t \leq b \), which has length\n\n\[ \mathcal{L}\left( {\mathbf{r}}_{0}\right) = {\int }_{a}^{b}\frac{2}{1 - {\left\lbrack x\left( t\right) \right\rbrack }^{2}}\sqrt{{\left( {x}^{\prime }\left( t\right) \right) }^{2}}{dt}. \]\n\nThe curve \( {\mathbf{r}}_{0} \) is essentially the shadow of \( \mathbf{r} \) on the real axis.\n\nOne can compare the integrands directly to see that \( \mathcal{L}\left( \mathbf{r}\right) \geq \mathcal{L}\left( {\mathbf{r}}_{0}\right) \) .\n\nSince transformations in \( \mathcal{H} \) preserve arc-length and hyperbolic lines, it follows that the shortest path between any two points in \( \mathbb{D} \) is along the hyperbolic line through them.
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Yes
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Corollary 5.3.10 The hyperbolic distance function is a metric on the hyperbolic plane. In particular, for any points \( p, q, u \) in \( \mathbb{D} \)\n\n1. \( {d}_{H}\left( {p, q}\right) \geq 0 \), and \( {d}_{H}\left( {p, q}\right) = 0 \) if and only if \( p = q \) ;\n\n2. \( {d}_{H}\left( {p, q}\right) = {d}_{H}\left( {q, p}\right) \) ; and\n\n3. \( {d}_{H}\left( {p, q}\right) + {d}_{H}\left( {q, u}\right) \geq {d}_{H}\left( {p, u}\right) \) .
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Proof. Recall our formula for the hyperbolic distance between two points in\n\nTheorem 5.3.3:\n\[ \n{d}_{H}\left( {p, q}\right) = \ln \left\lbrack \frac{\left| {1 - \bar{p}q}\right| + \left| {q - p}\right| }{\left| {1 - \bar{p}q}\right| - \left| {q - p}\right| }\right\rbrack .\n\]\n\nThis expression is always non-negative because the quotient inside the natural log is always greater than or equal to 1 . In fact, the expression equals 1 (so that the distance equals 0) if and only if \( p = q \) .\n\nNote further that this formula is symmetric. Interchanging \( p \) and \( q \) leaves the distance unchanged.\n\nFinally, the hyperbolic distance formula satisfies the triangle inequality because hyperbolic lines are geodesics.
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Yes
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Two paths from \( p = {.5i} \) to \( q = {.5} + {.5i} \) are shown below: the (solid) hyperbolic segment from \( p \) to \( q \), and the (dashed) path \( \mathbf{r} \) that looks like a Euclidean segment. Which path is shorter?
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We may compute the length of the hyperbolic segment connecting \( p \) and \( q \) with the distance formula from Theorem 5.3.3. This distance is approximately 1.49 units.\n\nBy contrast, consider the path in \( \mathbb{D} \) corresponding to the Euclidean line segment from \( p \) to \( q \) . This path may be described by \( \mathbf{r}\left( t\right) = t + \frac{1}{2}i \) for \( 0 \leq t \leq \frac{1}{2} \) . Then \( {\mathbf{r}}^{\prime }\left( t\right) = 1 \) and\n\n\[ \mathcal{L}\left( \mathbf{r}\right) = {\int }_{0}^{\frac{1}{2}}\frac{2}{1 - \left( {{t}^{2} + \frac{1}{4}}\right) }{dt} \]\n\n\[ \approx {1.52}\text{.} \]\n\nIt is no surprise that the hyperbolic segment connecting \( p \) to \( q \) is a shorter path in \( \left( {\mathbb{D},\mathcal{H}}\right) \) than the Euclidean line segment connecting them.
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Yes
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The area of a circle in \( \left( {\mathbb{D},\mathcal{H}}\right) \) .
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Suppose our region is given by a circle whose hyperbolic radius is \( a \) . Since area is an invariant, we may as well assume the circle is centered at the origin. Let \( x \) be the point at which the circle intersects the positive real axis (so \( 0 < x < 1 \) ), as pictured below. Then, by the distance formula\n\n\[ a = \ln \left( \frac{1 + x}{1 - x}\right) \]\n\nSolving for \( x \), we have\n\n\[ x = \frac{{e}^{a} - 1}{{e}^{a} + 1} \]\n\nThis circular region may be described in polar coordinates by \( 0 \leq \theta \leq \) \( {2\pi } \) and \( 0 \leq r \leq x \) . The area of the region is then given by the following integral, which we compute with the \( u \) -substitution \( u = 1 - {r}^{2} \) :\n\n\[ {\int }_{0}^{2\pi }{\int }_{0}^{x}\frac{4r}{{\left( 1 - {r}^{2}\right) }^{2}}{drd\theta } \]\n\n\[ = \left( {{\int }_{0}^{2\pi }{d\theta }}\right) \left( {{\int }_{1}^{1 - {x}^{2}}\frac{-2}{{u}^{2}}{du}}\right) \]\n\n\[ = {2\pi }\left\lbrack {\left. \frac{2}{u}\right| }_{1}^{1 - {x}^{2}}\right\rbrack \]\n\n\[ = {2\pi }\left\lbrack {\frac{2}{1 - {x}^{2}} - 2}\right\rbrack \]\n\n\[ = {4\pi }\frac{{x}^{2}}{1 - {x}^{2}} \]\n\nReplace \( x \) in terms of \( a \) to obtain\n\n\[ {4\pi }\frac{{\left( {e}^{a} - 1\right) }^{2}}{{\left( {e}^{a} + 1\right) }^{2}} \cdot \frac{{\left( {e}^{a} + 1\right) }^{2}}{{\left( {e}^{a} + 1\right) }^{2} - {\left( {e}^{a} - 1\right) }^{2}} = {4\pi }\frac{{\left( {e}^{a} - 1\right) }^{2}}{4{e}^{a}} \]\n\n\[ = {4\pi }{\left( \frac{{e}^{a} - 1}{2{e}^{a/2}}\right) }^{2} \]\n\n\[ = {4\pi }{\left( \frac{{e}^{a/2} - {e}^{-a/2}}{2}\right) }^{2}. \]\n\nThis last expression can be rewritten using the hyperbolic sine function, evaluated at \( a/2 \) .
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Yes
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Theorem 5.4.8 Any ideal triangle has area equal to \( \pi \) .
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Proof. Since all ideal triangles are congruent, assume our triangle \( \Delta \) is the ideal triangle shown in Figure 5.4.7.\n\nBut then \( \Delta \) can be partitioned into two \( \frac{2}{3} \) -ideal triangles by drawing the vertical hyperbolic line from 0 along the imaginary axis to ideal point \( i \) . Each \( \frac{2}{3} \) -ideal triangle has interior angle \( \pi /2 \), so \( \Delta \) has area \( \pi /2 + \pi /2 = \pi \) .
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Yes
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Theorem 5.4.9 The area of a hyperbolic triangle in \( \left( {\mathbb{D},\mathcal{H}}\right) \) having interior angles \( \alpha ,\beta \), and \( \gamma \) is\n\n\[ A = \pi - \left( {\alpha + \beta + \gamma }\right) \]
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Proof. Consider Figure 5.4.10 containing triangle \( {\Delta pqr} \) . We have extended segment \( {qp} \) to the ideal point \( t, u \) is an ideal point of line \( {rq} \), and \( v \) is an ideal point of line \( {pr} \) . The area of the ideal triangle \( {\Delta tuv} \) is \( \pi \) . Notice that regions \( {R}_{1} \) , \( {R}_{2} \), and \( {R}_{3} \) are all \( \frac{2}{3} \) -ideal triangles contained within the ideal triangle. Consider \( {R}_{1} \), whose ideal points are \( u \) and \( t \), and whose interior angle is \( \angle {uqt} \) . Since the line through \( q \) and \( r \) has ideal point \( u \), the interior angle of \( {R}_{1} \) is \( \angle {uqt} = \pi - \beta \) . Similarly, \( {R}_{2} \) has interior angle \( \pi - \alpha \) and \( {R}_{3} \) has interior angle \( \pi - \gamma \) .\n\nLet \( R \) denote the triangle region \( \Delta {pqr} \) whose area \( A\left( R\right) \) we want to compute. We then have the following relationships among areas:\n\n\[ \pi = A\left( R\right) + A\left( {R}_{1}\right) + A\left( {R}_{2}\right) + A\left( {R}_{3}\right) \]\n\n\[ = A\left( R\right) + \left\lbrack {\pi - \left( {\pi - \alpha }\right) }\right\rbrack + \left\lbrack {\pi - \left( {\pi - \beta }\right) }\right\rbrack + \left\lbrack {\pi - \left( {\pi - \gamma }\right) }\right\rbrack . \]\n\nSolving for \( A\left( R\right) \) ,\n\n\[ A\left( R\right) = \pi - \left( {\alpha + \beta + \gamma }\right) \]\n\nand this completes the proof.
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Yes
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Corollary 5.4.13 Hyperbolic hypotenuse theorem. In a right hyperbolic triangle with hyperbolic side lengths \( a \) and \( b \), and hypotenuse \( c \) ,
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\[ \cosh \left( c\right) = \cosh \left( a\right) \cosh \left( b\right) . \]
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Yes
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Suppose a two-dimensional ship is plopped down in \( \mathbb{D} \). What would the pilot see? How would the ship move? How would the pilot describe the world? Are all points equivalent in this world? Could the pilot figure out whether the universe adheres to hyperbolic geometry as opposed to, say, Euclidean geometry?
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Recall what we know about hyperbolic geometry. First of all, any two points in the hyperbolic plane are congruent, so the geometry is homogeneous. The pilot could not distinguish between any two points, geometrically.\n\nSecond, the shortest path between two points is the hyperbolic line between them, so light would travel along these hyperbolic lines, assuming light follows geodesics. The pilot's line of sight would follow along these lines, and the ship would move along these lines to fly as quickly as possible from \( p \) to \( q \), assuming no pesky asteroid fields block the path. To observe a galaxy at point \( q \) from the point \( p \) (as in the diagram below), the pilot would point a telescope in the direction of line \( L \), the line along which the light from the galaxy travels to reach the telescope.\n\nWith a well-defined metric, we can say more. The pilot will view the hyperbolic plane as infinite and without boundary. In theory, the pilot can make an orbit of arbitrary radius about an asteroid located anywhere in the space.\n\nTo test for hyperbolic geometry, perhaps the pilot can turn to triangles. The angles of a triangle in the hyperbolic plane sum to less than \( {180}^{ \circ } \), but only noticeably so for large enough triangles. In our disk model of hyperbolic geometry, we can easily observe this angle deficiency. In the figure below, triangle \( {\Delta zuw} \) has angle sum about \( {130}^{ \circ } \), and \( {\Delta pqr} \) has angle sum of about \( {22}^{ \circ } \) ! Whether an intrepid 2-D explorer could map out such a large triangle depends on how much ground she could cover relative to the size of her universe. We will have more to say about such things in Chapters 7 and 8.
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No
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Inscribe a circle in an ideal triangle.
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We show that if one inscribes a circle in any ideal triangle, its points of tangency form an equilateral triangle with side lengths equal to \( 2\ln \left( \varphi \right) \) where \( \varphi \) is the golden ratio \( \left( {1 + \sqrt{5}}\right) /2 \) .\n\nSince all ideal triangles are congruent, we choose one that is convenient to work with. Consider the ideal triangle with ideal points -1,1, and \( i \) .\n\nThe hyperbolic line \( {L}_{1} \) joining -1 and \( i \) is part of the circle \( {C}_{1} \) with radius 1 centered at \( - 1 + i \) . The hyperbolic line \( {L}_{2} \) joining \( i \) and 1 is part of the circle \( {C}_{2} \) with radius 1 centered at \( 1 + i \) . Let \( C \) denote the circle with radius 2 centered at \( - 1 + {2i} \), as pictured below.\n\n\n\nInversion about \( C \) gives a hyperbolic reflection of \( \mathbb{D} \) that maps \( {L}_{1} \) onto the real axis. Indeed, the circle \( {C}_{1} \), since it passes through the center of \( C \), gets mapped to a line - the real axis, in fact. Moreover, hyperbolic reflection across the imaginary axis maps \( {L}_{1} \) onto \( {L}_{2} \) . Let \( c \) be the point of intersection of these two hyperbolic lines of reflection, as pictured. The hyperbolic circle with hyperbolic center \( c \) that passes through the origin will be tangent to the real axis, and will thus inscribe the ideal triangle. Let the points of tangency on \( {L}_{1} \) and \( {L}_{2} \) be \( p \) and \( q \), respectively.\n\nThe point \( q \) can be found analytically as the point of intersection of circles \( C \) and \( {C}_{2} \) . Working it out, one finds \( q = \frac{1}{5} + \frac{2}{5}i \) . Thus,\n\n\[{d}_{H}\left( {0, q}\right) = \ln \left( \frac{1 + \left| q\right| }{1 - \left| q\right| }\right)\]\n\n\[= \ln \left( \frac{1 + \frac{1}{\sqrt{5}}}{1 - \frac{1}{\sqrt{5}}}\right)\]\n\n\[= \ln \left( \frac{\sqrt{5} + 1}{\sqrt{5} - 1}\right)\]\n\n\[= \ln \left( \frac{{\left( \sqrt{5} + 1\right) }^{2}}{\left( {\sqrt{5} - 1}\right) \left( {\sqrt{5} + 1}\right) }\right)\]\n\n\[= \ln \left( \frac{{\left( \sqrt{5} + 1\right) }^{2}}{4}\right)\]\n\n\[= \ln \left( {\varphi }^{2}\right)\]\n\n\[= 2\ln \left( \varphi \right) \text{.}
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Yes
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Example 5.5.4 The distance between \( {ri} \) and \( {si} \) .
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For \( r > s > 0 \) we compute the distance between \( {ri} \) and \( {si} \) in the upper half-plane model.\n\nThe hyperbolic line through \( {ri} \) and \( {si} \) is the positive imaginary axis, having ideal points 0 and \( \infty \) . Thus,\n\n\[ {d}_{U}\left( {{ri},{si}}\right) = \ln \left( \left( {{ri},{si};0,\infty }\right) \right) \]\n\n\[ = \frac{{ri} - 0}{{ri} - \infty } \cdot \frac{{si} - \infty }{{si} - 0} \]\n\n\[ = \ln \left( \frac{r}{s}\right) \]
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Yes
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The area of a \( \frac{2}{3} \) -ideal triangle.
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Suppose \( w \in \mathbb{U} \) is on the unit circle, and consider the \( \frac{2}{3} \) -ideal triangle \( {1w}\infty \) as pictured.\n\nIn particular, suppose the interior angle at \( w \) is \( \alpha \), so that \( w = {e}^{i\left( {\pi - \alpha }\right) } \) where \( 0 < \alpha < \pi \).\n\nThe area of this \( \frac{2}{3} \) -ideal triangle is thus\n\n\[ A = {\int }_{\cos \left( {\pi - \alpha }\right) }^{1}{\int }_{\sqrt{1 - {x}^{2}}}^{\infty }\frac{1}{{y}^{2}}{dydx} \]\n\n\[ = {\int }_{\cos \left( {\pi - \alpha }\right) }^{1}\frac{1}{\sqrt{1 - {x}^{2}}}{dx}. \]\n\nWith the trig substituion \( \cos \left( \theta \right) = x \), so that \( \sqrt{1 - {x}^{2}} = \sin \left( \theta \right) \) and \( - \sin \left( \theta \right) {d\theta } = {dx} \), the integral becomes\n\n\[ = {\int }_{\pi - \alpha }^{0}\frac{-\sin \left( \theta \right) }{\sin \left( \theta \right) }{d\theta } \]\n\n\[ = \pi - \alpha \text{.} \]
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Yes
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Lemma 6.1.2 Given two diametrically opposed points on the unit sphere, their image points under stereographic projection are antipodal points in \( {\mathbb{C}}^{ + } \) .
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Proof. First note that the north pole \( N = \left( {0,0,1}\right) \) and the south pole \( S = \left( {0,0, - 1}\right) \) are diametrically opposed points and they get sent by \( \phi \) to \( \infty \) and 0, respectively, in \( {\mathbb{C}}^{ + } \) ; so the lemma holds in this case.\n\nNow suppose \( P = \left( {a, b, c}\right) \) is a point on the sphere with \( \left| c\right| \neq 1 \), and \( Q = \) \( \left( {-a, - b, - c}\right) \) is diametrically opposed to \( P \) . The images of these two points under stereographic projection are\n\n\[ \phi \left( P\right) = \frac{a}{1 - c} + \frac{b}{1 - c}i\text{ and }\phi \left( Q\right) = \frac{-a}{1 + c} - \frac{b}{1 + c}i. \]\n\nIf we expand the following product\n\n\[ \phi \left( P\right) \cdot \overline{\phi \left( Q\right) } = \left\lbrack {\frac{a}{1 - c} + \frac{b}{1 - c}i}\right\rbrack \cdot \left\lbrack {\frac{-a}{1 + c} + \frac{b}{1 + c}i}\right\rbrack , \]\n\nwe obtain\n\n\[ \phi \left( P\right) \cdot \overline{\phi \left( Q\right) } = - \frac{{a}^{2} + {b}^{2}}{1 - {c}^{2}} \]\n\nwhich reduces to -1 since \( {a}^{2} + {b}^{2} + {c}^{2} = 1 \) .\n\nThus, diametrically opposed points on the sphere get mapped via stereographic projection to antipodal points in the extended plane.
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Yes
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Lemma 6.1.4 If a cline in \( {\mathbb{C}}^{ + } \) contains two antipodal points then it is a great circle.
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Proof. Suppose \( C \) is a cline in \( {\mathbb{C}}^{ + } \) containing antipodal points \( p \) and \( {p}_{a} \), and suppose \( q \) is any other point on \( C \) . We show \( {q}_{a} \) is also on \( C \) .\n\nIf \( C \) is a line, it must go through the origin since \( p \) and \( {p}_{a} \) are on the same line through the origin. Since \( q \) and \( {q}_{a} \) are also on the same line through the origin, if \( q \) is on \( C \) then \( {q}_{a} \) is too.\n\nIf \( C \) is a circle, then the origin of the plane is in the interior of \( C \) and the chord \( p{p}_{a} \) contains the origin, as pictured in Figure 6.1.5. The line through \( q \) and the origin intersects \( C \) at another point, say \( w \) . We show \( w = {q}_{a} \) .\n\n\n\nFigure 6.1.5 A cline \( C \) containing a pair of antipodal points \( p \) and \( {p}_{a} \) must be a great cirlce. In the figure, \( w = {q}_{a} \) .\n\nThe intersecting chords theorem (Book III, Proposition 35 of Euclid's Elements) applied to this figure tells us that \( \left| p\right| \cdot \left| {p}_{a}\right| = \left| q\right| \cdot \left| w\right| \) . (We leave the proof to Exercise 6.1.7. The proof essentially follows that of Lemma 3.2.7, except we consider a point inside the given circle.) As antipodal points, \( \left| p\right| \cdot \left| {p}_{a}\right| = 1 \), and it follows that \( \left| w\right| = 1/\left| q\right| \) . Since the segment \( {wq} \) contains the origin, it follows that \( w \) is the point antipodal to \( q \) .
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No
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Theorem 6.1.6 Reflection of \( {\mathbb{S}}^{2} \) about a great circle corresponds via stereographic projection to inversion about a great circle in \( {\mathbb{C}}^{ + } \) .
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We work through the relationship in one case, and refer the interested reader to \( \left\lbrack {10}\right\rbrack \) for the general proof.
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No
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We argue that reflection of \( {\mathbb{S}}^{2} \) about the equator corresponds to inversion about the unit circle.
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First of all, stereographic projection sends the equator of \( {\mathbb{S}}^{2} \) to the unit circle in \( {\mathbb{C}}^{ + } \) . Now, reflection of \( {\mathbb{S}}^{2} \) across the equator sends the point \( P = \left( {a, b, c}\right) \) to the point \( {P}^{ * } = \left( {a, b, - c}\right) \) . We must argue that \( \phi \left( P\right) \) and \( \phi \left( {P}^{ * }\right) \) are symmetric with respect to the unit circle.\n\nIf \( \left| c\right| = 1 \) then \( P \) and \( {P}^{ * } \) correspond to the north and south poles, and their image points are 0 and \( \infty \), and these points are symmetric with respect to the unit circle. So, assume \( \left| c\right| \neq 1 \) . Notice that\n\n\[ \phi \left( P\right) = \frac{a}{1 - c} + \frac{b}{1 - c}i\text{ and }\phi \left( {P}^{ * }\right) = \frac{a}{1 + c} + \frac{b}{1 + c}i \]\n\nare on the same ray beginning at the origin. Indeed, one is the positive scalar multiple of the other:\n\n\[ \phi \left( {P}^{ * }\right) = \frac{1 - c}{1 + c}\phi \left( P\right) \]\n\nMoreover,\n\n\[ \left| {\phi \left( P\right) }\right| \cdot \left| {\phi \left( {P}^{ * }\right) }\right| = \frac{1 - c}{1 + c} \cdot {\left| \phi \left( P\right) \right| }^{2} \]\n\n\[ = \frac{1 - c}{1 + c} \cdot \frac{{a}^{2} + {b}^{2}}{{\left( 1 - c\right) }^{2}} \]\n\n\[ = \frac{{a}^{2} + {b}^{2}}{1 - {c}^{2}} \]\n\n\[ = 1\text{.} \]\n\nAgain, the last equality holds because \( {a}^{2} + {b}^{2} + {c}^{2} = 1 \) . Thus, \( \phi \left( P\right) \) and \( \phi \left( {P}^{ * }\right) \) are symmetric with respect to the unit circle. It follows that inversion in the unit circle, \( {i}_{{\mathbb{S}}^{1}} \), corresponds to reflection of \( {\mathbb{S}}^{2} \) across the equator, call this map \( R \), by the equation\n\n\[ {i}_{{\mathbb{S}}^{1}} \circ \phi = \phi \circ R \]
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Yes
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Theorem 6.1.8 The image of a circle on \( {\mathbb{S}}^{2} \) via stereographic projection is a cline in \( {\mathbb{C}}^{ + } \) . Moreover, the pre-image of a circle in \( {\mathbb{C}}^{ + } \) is a circle on \( {\mathbb{S}}^{2} \) . The pre-image of a line in \( {\mathbb{C}}^{ + } \) is a circle on \( {\mathbb{S}}^{2} \) that goes through \( N = \left( {0,0,1}\right) \) .
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In fact, one can offer a constructive proof of this theorem. A circle on \( {\mathbb{S}}^{2} \) can be represented as the intersection of \( {\mathbb{S}}^{2} \) with a plane \( {Ax} + {By} + {Cz} + D = 0 \) in 3-dimensional space. One can show that the circle in \( {\mathbb{S}}^{2} \) defined by \( {Ax} + {By} + {Cz} + \) \( D = 0 \) gets mapped by \( \phi \) to the cline \( \left( {C + D}\right) \left( {{u}^{2} + {v}^{2}}\right) + {2Au} + {2Bv} + \left( {D - C}\right) = 0 \) in the plane (described via \( u, v \) cartesian coordinates); conversely the circle \( \left| {w - {w}_{0}}\right| = r \) in \( \mathbb{C} \) gets mapped by the inverse function \( {\phi }^{-1} \) to the circle in \( {\mathbb{S}}^{2} \) given by the plane \( - 2{x}_{0}x - 2{y}_{o}y + \left( {1 - {\left| {w}_{0}\right| }^{2} + {r}^{2}}\right) z + \left( {1 + {\left| {w}_{0}\right| }^{2} - {r}^{2}}\right) = 0 \) where \( {w}_{0} = {x}_{0} + {y}_{0}i \)
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Yes
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Example 6.1.9 The image of a circle under \( \phi \) .\n\nConsider the circle on \( {\mathbb{S}}^{2} \) defined by the vertical plane \( x = - \frac{1}{2} \) . In standard form, this plane has constants \( A = 2, B = C = 0 \), and \( D = 1 \), so the image under \( \phi \) is the circle\n\n\[ \left( {{u}^{2} + {v}^{2}}\right) + {4u} + 1 = 0. \]
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Completing the square we obtain the circle\n\n\[ {\left( u + 2\right) }^{2} + {v}^{2} = 3 \]\n\nhaving center \( \left( {-2,0}\right) \) and radius \( \sqrt{3} \).
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Yes
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Constructing an antipodal point. Suppose \( z \) is a point inside the unit circle. Prove that the following construction, which is depicted in Figure 6.1.11, gives \( {z}_{a} \), the point antipodal to \( z \) : (1) Draw the line through \( z \) and the origin; (2) draw the line through the origin perpendicular to line (1), and let \( T \) be on line (2) and the unit circle; (3) construct the segment \( {zT} \) ; (4) construct the perpendicular to segment (3) at point \( T \) . Line (4) intersects line (1) at the point \( {z}_{a} \) . Use similar triangles to prove that \( {z}_{a} = - \frac{1}{{\left| z\right| }^{2}}z \) .
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\n\nFigure 6.1.11 Constructing the antipodal point to \( z \) .
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No
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Theorem 6.2.4 If a Möbius transformation preserves antipodal points, then it is an elliptic Möbius transformation.
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Proof. Suppose \( T \) is a Möbius transformation that preserves antipodal points. If \( T \) is not the identity map then it must fix one or two points. However, if \( T \) preserves antipodal points and fixes a point \( p \), then it must fix its antipodal point \( {p}_{a} \) . Thus, \( T \) must have two fixed points and \( T \) has normal form\n\n\[ \frac{T\left( z\right) - p}{T\left( z\right) - {p}_{a}} = \lambda \frac{z - p}{z - {p}_{a}} \]\n\nTo show \( T \) is an elliptic Möbius transformation, we must show that \( \left| \lambda \right| = 1 \) . Setting \( z = 0 \) the normal form reduces to\n\n\[ \frac{T\left( 0\right) - p}{T\left( 0\right) - {p}_{a}} = \lambda \frac{p}{{p}_{a}} \]\n\nSolve for \( T\left( 0\right) \) to obtain\n\n\[ T\left( 0\right) = \frac{p{p}_{a} - {\lambda p}{p}_{a}}{{p}_{a} - {\lambda p}}. \]\n\nOn the other hand, setting \( z = \infty \) and solving for \( T\left( \infty \right) \), one checks that\n\n\[ T\left( \infty \right) = \frac{p - \lambda {p}_{a}}{1 - \lambda } \]\n\nSince \( T \) preserves antipodal points, \( T\left( 0\right) \cdot \overline{T\left( \infty \right) } = - 1 \) ; therefore, we have\n\n\[ \frac{p{p}_{a} - {\lambda p}{p}_{a}}{{p}_{a} - {\lambda p}} \cdot \frac{\bar{p} - \bar{\lambda }\overline{{p}_{a}}}{1 - \bar{\lambda }} = - 1. \]\n\nIf we expand this expression and solve it for \( \lambda \bar{\lambda } \) (using the fact that \( p \cdot \overline{{p}_{a}} = - 1 \) , and \( \left. {p \neq {p}_{a}}\right) \), we obtain \( \lambda \bar{\lambda } = 1 \), from which it follows that \( \left| \lambda \right| = 1 \) . Thus \( T \) is an elliptic Möbius transformation.
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Yes
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Theorem 6.2.11 There is a unique elliptic line connecting two points \( p \) and \( q \) in \( {\mathbb{P}}^{2} \) .
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Proof. Suppose \( p \) and \( q \) are distinct points in \( {\mathbb{P}}^{2} \) . This means \( q \neq {p}_{a} \) as points in \( {\mathbb{C}}^{ + } \) . Construct the antipodal point \( {p}_{a} \), which gives us three distinct points in \( {\mathbb{C}}^{ + } : p, q \) and \( {p}_{a} \) . There exists a unique cline through these three points. Since this cline goes through \( p \) and \( {p}_{a} \), it is an elliptic line by Lemma 6.1.4.
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Yes
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Theorem 6.2.12 The set of elliptic lines is a minimally invariant set of elliptic geometry.
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Proof. By definition, any transformation \( T \) in \( \mathcal{S} \) preserves antipodal points. Thus, if \( L \) is an elliptic line, then \( T\left( L\right) \) is as well, and the set of elliptic lines is an invariant set of elliptic geometry.\n\nTo show the set is minimally invariant, we appeal to Theorem 4.1.10, and prove that any two elliptic lines are congruent. To see this, notice that any elliptic line \( L \) is congruent to the elliptic line on the real axis. Indeed, for any points \( {z}_{0} \) and \( {z}_{1} \) on \( L \), Lemma 6.2.3 ensures the existence of a transformation \( T \) in \( \mathcal{S} \) that sends \( {z}_{0} \) to the origin, and \( {z}_{1} \) to the real axis. It follows that \( T\left( L\right) \) is the real axis. Since all elliptic lines are congruent to the real axis, any two elliptic lines are congruent.
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Yes
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Theorem 6.2.13 Any two elliptic lines intersect in \( {\mathbb{P}}^{2} \) .
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Proof. Given any two elliptic lines, apply a transformation \( T \) in \( \mathcal{S} \) that sends one of them to the real axis. It is enough to prove that any elliptic line in \( {\mathbb{P}}^{2} \) must intersect the real axis. Suppose \( M \) is an arbitrary elliptic line in \( {\mathbb{P}}^{2} \) and \( z \) is a point on \( M \) . If \( \operatorname{Im}\left( z\right) = 0 \) then \( z \) is on the real axis and we are done.\n\nIf \( \operatorname{Im}\left( z\right) > 0 \), then \( z \) lies above the real axis. It follows that \( \operatorname{Im}\left( {z}_{a}\right) < 0 \) by the definition of the antipodal point \( {z}_{a} \) . Since \( M \) contains both \( z \) and \( {z}_{a} \) it intersects the real axis at some point.\n\nIf \( \operatorname{Im}\left( z\right) < 0 \), then \( \operatorname{Im}\left( {z}_{a}\right) > 0 \) and \( M \) must intersect the real axis as before. In either case, \( M \) must intersect the real axis, and it follows that any two elliptic lines must intersect.
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Yes
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Theorem 6.3.4 The distance between two points \( p \) and \( q \) in \( \left( {{\mathbb{P}}^{2},\mathcal{S}}\right) \) is\n\n\[ \n{d}_{S}\left( {p, q}\right) = \min \left\{ {2\arctan \left( \left| \frac{q - p}{1 + \bar{p}q}\right| \right) ,2\arctan \left( \left| \frac{1 + \bar{p}q}{q - p}\right| \right) }\right\} .\n\]
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Proof. We first determine the elliptic distance between the origin and a point \( x \) (with \( 0 < x \leq 1 \) ) on the positive real axis.\n\nThe elliptic line through 0 and \( x \) lives on the real axis, and we may parameterize the \
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No
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Lemma 6.3.9 The area of a lune. In \( \left( {{\mathbb{P}}^{2},\mathcal{S}}\right) \), the area of a lune with angle \( \alpha \) is \( {2\alpha } \) .
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Proof. To compute the area of a lune, first move the vertex of the lune to the origin in such a way that one leg of the lune lies on the real axis, as in Figure 6.3.10. Then half of the lunar region can be described in polar coordinates by \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq \alpha \ ).\n\n\n\nFigure 6.3.10 A lune whose elliptic lines intersect at the origin.\n\nSo the area of a lune having angle \( \alpha \) is given by\n\n\[ A = 2{\int }_{0}^{\alpha }{\int }_{0}^{1}\frac{4r}{{\left( 1 + {r}^{2}\right) }^{2}}{drd\theta }\ ]\n\n\[ \text{(Let}u = 1 + {r}^{2}\text{)} \]\n\n\[ = 2{\int }_{0}^{\alpha }\left\lbrack {\left. \frac{-2}{1 + {r}^{2}}\right| }_{0}^{1}\right\rbrack {d\theta } \]\n\n\[ = 2{\int }_{0}^{\alpha }{1d\theta } \]\n\n\[ = {2\alpha }\text{.} \]\n
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Yes
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Example 6.3.11 Triangle area in \( \left( {{\mathbb{P}}^{2},\mathcal{S}}\right) \) .
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Triangle \( {\Delta pqr} \) below is formed from 3 elliptic lines. Notice that each corner of the triangle determines a lune, and that the three lunes cover the entire projective plane, with some overlap. In particular, the three lunes in sum cover the triangle three times, so the sum of the three lune areas equals the area of the entire projective plane plus two times the area of the triangle.\n\n\n\nThe area of the entire projective plane is \( {2\pi } \) (see Exercise 6.3.9), so we have the following relation:\n\n\[ {2\pi } + 2 \cdot A\left( {\Delta pqr}\right) = A\left( {\text{ Lune }p}\right) + A\left( {\text{ Lune }q}\right) + A\left( {\text{ Lune }r}\right) \]\n\n\[ {2\pi } + 2 \cdot A\left( {\Delta pqr}\right) = \left( {{2\alpha } + {2\beta } + {2\gamma }}\right) \]\n\n\[ A\left( {\Delta pqr}\right) = \left( {\alpha + \beta + \gamma }\right) - \pi . \]
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Yes
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Theorem 6.3.12 In elliptic geometry \( \left( {{\mathbb{P}}^{2},\mathcal{S}}\right) \), the area of a triangle with angles \( \alpha ,\beta ,\gamma \) is\n\n\[ A = \left( {\alpha + \beta + \gamma }\right) - \pi . \]
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From this theorem it follows that the angles of any triangle in elliptic geometry sum to more than \( {180}^{ \circ } \) .
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No
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Corollary 6.3.16 Elliptic hypotenuse theorem. In a right triangle in \( \\left( {{\\mathbb{P}}^{2},\\mathcal{S}}\\right) \) with elliptic side lengths \( a \) and \( b \), and hypotenuse \( c \) ,
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\[ \\cos \\left( c\\right) = \\cos \\left( a\\right) \\cos \\left( b\\right) \]
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Yes
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Lemma 7.2.1 Assume \( k > 0 \) . A lune in \( \left( {{\mathbb{P}}_{k}^{2},{\mathcal{S}}_{k}}\right) \) with interior angle \( \alpha \) has area \( {2\alpha }/k \) .
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Proof. Without loss of generality, we may consider the vertex of our lune to be the origin. As before, elliptic lines through the origin must also pass through \( \infty \), so our two lines forming the lune are Euclidean lines. After a convenient rotation, we may further assume one of these lines is the real axis, so that the lune resembles the one in Figure 6.3.10. To compute the area of the lune, compute the integral\n\n\[ A = 2{\int }_{0}^{\alpha }{\int }_{0}^{1/\sqrt{k}}\frac{4r}{{\left( 1 + k{r}^{2}\right) }^{2}}{drd\theta }.\]\n\nLetting \( u = 1 + k{r}^{2} \) so that \( {du} = {2krdr} \), the bounds of integration change from \( \left\lbrack {0,1/\sqrt{k}}\right\rbrack \) to \( \left\lbrack {1,2}\right\rbrack \) . Then,\n\n\[ A = 2{\int }_{0}^{\alpha }\frac{2}{k}{\int }_{1}^{2}\frac{du}{{u}^{2}}{d\theta } = \frac{4}{k}{\int }_{0}^{\alpha }\frac{1}{2}{d\theta } = \frac{2\alpha }{k}.\]\n\nThus, the angle of a lune with interior angle \( \alpha \) is \( {2\alpha }/k \).
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Yes
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Lemma 7.2.2 In elliptic geometry with curvature \( k \), the area of a triangle with angles \( \alpha ,\beta \), and \( \gamma \) is\n\n\[ A = \frac{1}{k}\left( {\alpha + \beta + \gamma - \pi }\right) \]
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Proof. As in the case \( k = 1 \), the area of any triangle may be determined from the area of three lunes and the total area of \( {\mathbb{P}}_{k}^{2} \), as depicted in Example 6.3.11.
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No
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Theorem 7.3.3 Lobatchevsky's formula. In hyperbolic geometry with curvature \( k \), the hyperbolic distance \( d \) of a point \( z \) to a hyperbolic line \( L \) is related to the angle of parallelism \( \theta \) by the formula\n\n\[ \tan \left( {\theta /2}\right) = {e}^{-\sqrt{\left| k\right| }d}. \]
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Proof. For this proof, let \( s = \frac{1}{\sqrt{\left| k\right| }} \) . Note that \( s \) is the Euclidean radius of the circle at infinity in the disk model for hyperbolic geometry with curvature \( k \) . Since angles and lines and distances are preserved, assume \( z \) is the origin and \( L \) is orthogonal to the positive real axis, intersecting it at the point \( x \) (with \( 0 < x < s \) ).\n\n\n\nFigure 7.3.4 Deriving Lobatchevsky's formula.\n\nRecall the half-angle formula\n\n\[ \tan \left( {\theta /2}\right) = \frac{\tan \left( \theta \right) }{\sec \left( \theta \right) + 1}. \]\n\nAccording to Figure 7.3.4, \( \tan \left( \theta \right) = r/s \), where \( r \) is the Euclidean radius of the circle containing the hyperbolic line \( L \) . Furthermore, \( \sec \left( \theta \right) = \left( {x + r}\right) /s \), so\n\n\[ \tan \left( {\theta /2}\right) = \frac{r}{x + r + s}. \]\n\n(1)\n\nWe may express \( x \) and \( r \) in terms of the hyperbolic distance \( d \) from 0 to \( x \) . In Exercise 7.3.2 we prove the hyperbolic distance from 0 to \( x \) in \( \left( {{\mathbb{D}}_{k},{\mathcal{H}}_{k}}\right) \) is\n\n\[ d = s\ln \left( \frac{s + x}{s - x}\right) \]\n\nso that\n\n\[ x = s \cdot \frac{{e}^{d/s} - 1}{{e}^{d/s} + 1} \]\n\nExpress \( r \) in terms of \( d \) by first expressing it in terms of \( x \) . Note that segment \( x{x}^{ * } \) is a diameter of the circle containing \( L \), where \( {x}^{ * } = \frac{-1}{kx} \) is the point symmetric to \( x \) with respect to the circle at infinity. Thus, \( r \) is half the distance from \( x \) to \( {x}^{ * } \):\n\n\[ r = - \frac{1 + k{x}^{2}}{2kx}. \]\n\nReplacing \( k \) with \( - 1/{s}^{2} \), we have\n\n\[ r = \frac{{s}^{2} - {x}^{2}}{2x}. \]\n\nOne checks that after writing \( x \) in terms of \( d, r \) is given by\n\n\[ r = \frac{{2s}{e}^{d/s}}{{e}^{{2d}/s} - 1}. \]\n\nSubstitute this expression for \( r \) into the equation labeled (1) in this proof, and after a dose of satisfactory simplifying one obtains the desired result:\n\n\[ \tan \left( {\theta /2}\right) = {e}^{-d/s}. \]\n\nSince \( s = 1/\sqrt{\left| k\right| } \) this completes the proof.
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Yes
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Theorem 7.4.2 For all real numbers \( k,\left( {{X}_{k},{G}_{k}}\right) \) is homogeneous and isotropic.
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Proof. Given any point \( p \) in \( {X}_{k} \), the transformation \( T\left( z\right) = \frac{z - p}{1 + k\bar{p}z} \) in \( {G}_{k} \) maps \( p \) to the origin. So all points in \( {X}_{k} \) are congruent to 0 . By the group structure of \( {G}_{k} \) it follows that any two points in \( {X}_{k} \) are congruent, so the geometry is homogeneous.\n\nTo show \( \left( {{X}_{k},{G}_{k}}\right) \) is isotropic we consider three cases.\n\nIf \( k < 0 \) then \( \left( {{X}_{k},{G}_{k}}\right) \) models hyperbolic geometry on the open disk with radius \( s = 1/\sqrt{\left| k\right| } \) . As such, \( {G}_{k} \) contains the sorts of Möbius transformations discussed in Chapter 5 and pictured in Figure 5.1.6. In particular, for any point \( p \in {\mathbb{D}}_{k},{G}_{k} \) contains all elliptic Möbius transformations that swirl points around type II clines of \( p \) and \( - 1/\left( {k\bar{p}}\right) \), the point symmetric to \( p \) with respect to the circle at infinity. These maps are preciley the rotations about the point \( p \) in this geometry: they rotate points in \( {X}_{k} \) around cicles centered at \( p \) .\n\nIf \( k = 0 \), then transformations in \( {G}_{0} \) have the form \( T\left( z\right) = {e}^{i\theta }\left( {z - {z}_{0}}\right) \) . Now, rotation by angle \( \theta \) about the point \( p \) in the Euclidean plane is given by \( T\left( z\right) = {e}^{i\theta }\left( {z - p}\right) + p \) . Setting \( {z}_{0} = p - p{e}^{-{i\theta }} \) we see that this rotation indeed lives in \( {G}_{0} \) .\n\nIf \( k > 0 \) then \( \left( {{X}_{k},{G}_{k}}\right) \) models elliptic geometry on the projective plane with radius \( s = 1/\sqrt{k} \) . As such, for each \( p \in {X}_{k},{G}_{k} \) contains all elliptic Möbius transformations that fix \( p \) and \( {p}_{a} \), the point antipodal to \( p \) with respect to the circle with radius \( s \) . Such a map rotates points around type II clines with respect to \( p \) and \( {p}_{a} \) . Since these type II clines correspond to circles in \( {X}_{k} \) centered at the fixed point, it follows that \( {G}_{k} \) contains all rotations.\n\nThus, for all \( k \in \mathbb{R},\left( {{X}_{k},{G}_{k}}\right) \) is isotropic.
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Yes
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Theorem 7.4.3 Suppose \( k \) is any real number, and we have a triangle in \( \left( {{X}_{k},{G}_{k}}\right) \) whose angles are \( \alpha ,\beta \), and \( \gamma \) and whose area is \( A \) . Then\n\n\[ \n{kA} = \left( {\alpha + \beta + \gamma - \pi }\right) .\n\]
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Proof of this tidy result has already appeared in pieces (see Exercise 1.2.2, Lemma 7.2.2, and Lemma 7.3.1); we emphasize that this triangle area formula reveals the locally Euclidean nature of all the geometries \( \left( {{X}_{k},{G}_{k}}\right) \) : a small triangle (one with area close to 0 ) will have an angle sum close to \( {180}^{ \circ } \) . Observe also that the closer \( \left| k\right| \) is to 0, the larger a triangle needs to be in order to detect an angle sum different from \( {180}^{ \circ } \) . Of course, if \( k = 0 \) the theorem tells us that the angles of a Euclidean triangle sum to \( \pi \) radians.
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No
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Theorem 7.4.4 Suppose a convex \( n \) -sided polygon \( \left( {n \geq 3}\right) \) in \( \left( {{X}_{k},{G}_{k}}\right) \) has interior angles \( {\alpha }_{i} \) for \( i = 1,2,\ldots, n \) . The area \( A \) of the \( n \) -gon is related to its interior angles by\n\n\[ \n{kA} = \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}}\right) - \left( {n - 2}\right) \pi .\n\]
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Proof. A convex \( n \) -gon can be divided into \( n - 2 \) triangles as in Figure 7.4.5. Observe that the area of the \( n \) -gon equals the sum of the areas of these triangles.\n\n\n\nFigure 7.4.5 Splitting an \( n \) -gon into \( n - 2 \) triangles in the case \( n = 8 \) .\n\nBy Theorem 7.4.3, the area \( {A}_{i} \) of the \( i \) th triangle \( {\Delta }_{i} \) is related to its angle sum by\n\n\[ \nk{A}_{i} = \left( {\sum \text{ angles in }{\Delta }_{i}}\right) - \pi .\n\]\n\nThus,\n\n\[ \n{kA} = \mathop{\sum }\limits_{{i = 1}}^{{n - 2}}k{A}_{i} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{i = 1}}^{{n - 2}}\left( {\sum \text{ angles in }{\Delta }_{i} - \pi }\right) .\n\]\n\nNow, the total angle sum of the \( n - 2 \) triangles equals the interior angle sum of the \( n \) -gon, so it follows that\n\n\[ \n{kA} = \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}}\right) - \left( {n - 2}\right) \pi .\n\]\n\nThis completes the proof.
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Yes
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Lemma 7.4.6 Suppose \( k \in \mathbb{R}, s = \frac{1}{\sqrt{\left| k\right| }} \) and \( 0 < x < s \) is a real number (if \( k = 0 \) , we just assume \( 0 < x \) ). In \( \left( {{X}_{k},{G}_{k}}\right) \), the circle centered at 0 through \( x \) has area\n\n\[ \frac{{4\pi }{x}^{2}}{1 + k{x}^{2}} \]
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Proof. Consider the circle centered at the origin that goes through the point \( x \) on the positive real axis, where \( 0 < x < s \) . The circular region matches the polar rectangle \( 0 < \theta < {2\pi } \) and \( 0 < r < x \), so the area is given by\n\n\[ {\int }_{0}^{2\pi }{\int }_{0}^{x}\frac{4r}{{\left( 1 + k{r}^{2}\right) }^{2}}{drd\theta } \]\n\nEvaluating this integral gives the result, and the details are left as an exercise.
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No
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Theorem 7.4.7 Unified Pythagorean Theorem. Suppose \( k \in \mathbb{R} \), and we have a geodesic right triangle in \( \left( {{X}_{k},{G}_{k}}\right) \) whose legs have length a and \( b \) and whose hypotenuse has length \( c \) . Then\n\n\[ A\left( c\right) = A\left( a\right) + A\left( b\right) - \frac{k}{2\pi }A\left( a\right) A\left( b\right) ,\]\n\nwhere \( A\left( r\right) \) denotes the area of a circle with radius \( r \) as measured in \( \left( {{X}_{k},{G}_{k}}\right) \) .
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Proof. Suppose \( k \in \mathbb{R} \) . If \( k = 0 \) the equation reduces to \( {c}^{2} = {a}^{2} + {b}^{2} \), which is true by the Pythagorean Theorem 1.2.1! Otherwise, assume \( k \neq 0 \) and let \( s = \frac{1}{\sqrt{\left| k\right| }} \), as usual. Without loss of generality we may assume our right triangle is defined by the points \( z = 0, p = x \), and \( q = {yi} \), where \( 0 < x, y < s \) . By construction, the legs \( {zp} \) and \( {zq} \) are Euclidean segments, and \( \angle {pzq} \) is right.\n\nLet \( a = {d}_{k}\left( {z, p}\right), b = {d}_{k}\left( {z, q}\right) \), and \( c = {d}_{k}\left( {p, q}\right) \) . By Lemma 7.4.6,\n\n\[ A\left( a\right) = \frac{{4\pi }{x}^{2}}{1 + k{x}^{2}},\text{ and }A\left( b\right) = \frac{{4\pi }{y}^{2}}{1 + k{y}^{2}}.\]\n\nTo find \( A\left( c\right) \), we first find \( {d}_{k}\left( {p, q}\right) \) . By invariance of distance,\n\n\[ {d}_{k}\left( {p, q}\right) = {d}_{k}\left( {0,\left| {T\left( q\right) }\right| }\right) ,\]\n\nwhere \( T\left( z\right) = \frac{z - p}{1 + k\bar{p}z} \) . Let \( t = \left| {T\left( q\right) }\right| = \frac{\left| yi - x\right| }{\left| 1 + kxyi\right| } \), which is a real number.\n\nNow, \( A\left( c\right) \) is the area of a circle with radius \( c \), and the circle centered at 0 through \( t \) has this radius, so\n\n\[ A\left( c\right) = \frac{{4\pi }{t}^{2}}{1 + k{t}^{2}} \]\n\nUsing the fact that \( {t}^{2} = \frac{{x}^{2} + {y}^{2}}{1 + {k}^{2}{x}^{2}{y}^{2}} \), one can now check by direct substitution that\n\n\[ A\left( c\right) = A\left( a\right) + A\left( b\right) - \frac{k}{2\pi }A\left( a\right) A\left( b\right) . \]
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Yes
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Let \( S \) consist of parallel planes in \( {\mathbb{R}}^{3} \). In particular, let \( S = \{ \left( {x, y, z}\right) \in \left. {{\mathbb{R}}^{3} \mid z = 0\text{or}z = 1}\right\} \) as in part (a) of the following diagram. Each point in \( S \) has a neighborhood of points that is an open 2-ball.
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Each point in \( S \) has a neighborhood of points that is an open 2-ball.
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No
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Let \( {\mathbb{T}}^{2} \) denote the torus surface in \( {\mathbb{R}}^{3} \) and \( {\mathbb{S}}^{2} \) denote the sphere, as usual. Figure 7.5.8 depicts two connected sums: \( {\mathbb{T}}^{2}\# {\mathbb{T}}^{2} \), and \( {\mathbb{S}}^{2}\# {\mathbb{T}}^{2} \). The surface \( {\mathbb{T}}^{2}\# {\mathbb{T}}^{2} \) is called the two-holed torus, and it is topologically equivalent to the surface labeled \( {\mathrm{H}}_{2} \) in Figure 7.5.10. One can shrink the length dimension of the connecting cylinder to make the one look like the other. The surface \( {\mathbb{S}}^{2}\# {\mathbb{T}}^{2} \) is homeomorphic to the torus \( {\mathbb{T}}^{2} \).
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To see this, observe that if one removes an open-2 ball from a sphere and attaches one end of a cylinder to the sphere along the boundary of the removed disk, the result is homeomorphic to a closed disk, as suggested in the following diagram. So, if one removes an open 2-ball from a surface \( X \) and then caps the hole with this sphere-with-cylinder shape, the net effect is patching the hole. In the arithmetic of connected sums, the sphere plays the role of 0 : \( {\mathbb{S}}^{2}\# X = X \) for any surface \( X \) .
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Yes
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Theorem 7.5.12 Any surface is homeomorphic to the sphere \( {\mathbb{S}}^{2} \), a handlebody surface \( {H}_{g} \) with \( g \geq 1 \), or a crosscap surface \( {C}_{g} \) with \( g \geq 1 \) . Moreover, no two surfaces in this list are homeomoprhic to each other.
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Two proofs of this theorem are floating around the literature now. The classic proof, which makes use of cell divisions, can be found, for instance, in [9]. The new proof, due to John Conway, bypasses the artificial constructs in the classic proof, and can be found in [12].
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No
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Why is the Klein bottle non-orientable?
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A bug leaving the screen on the right near the top would reappear on the left near the bottom. But take a closer look, the bug has become mirror-reversed. This orientation-reversing path exists because of a Möbius strip lurking in the Klein bottle. (Conisder, for instance, the thin horizontal strip formed by the dashed segments in the figure.)
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Yes
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Example 7.5.21 Attempted cell divisions of \( {H}_{1} \) .
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Three cell divisions of the torus are pictured below, along with one failed cell division. In each valid cell division, we count the number of faces, edges, and vertices of the cell division. To make an accurate count, one must take the edge identification into account.\n\n\n\nCell division (a) has two vertices, six edges, and four faces. One vertex is in the center of the rectangle, and the other vertex is in the corner (remember, all four corners get identified to a single point). As for the edges, four emanate from the center vertex, and we have two others: the horizontal edge along the boundary of the rectangle (appearing twice), and the vertical edge along the boundary (also appearing twice). Thus, the edges of the rectangle are also edges of the cell division, so the faces in this cell division are triangles, and there are four of them.\n\nCell division (b) has six vertices, eight edges, and two faces. There are four vertices in the interior of the rectangle, one vertex on the horizontal boundary of the rectangle, and one vertex on the vertical boundary. Notice that the corner point of the rectangle is not a vertex of the cell division. To count the edges, observe that four edges form the inner diamond, and one edge leaves each vertex of the diamond, for a total of eight edges. The boundary edges of the rectangle do not form edges in this cell division. Counting faces, we have one inside the diamond and one outside the diamond. Convince yourself that the region outside the diamond makes just one face.\n\nYou can check that cell division (c) has one vertex in the center of the rectangle, two edges (one is horizontal, the other is vertical), and one face.\n\nThe attempt (d) fails to be a cell division of the torus. Why is this? At first glance, we have four vertices, four edges, and two faces. The trouble here is the \
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Yes
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Theorem 7.5.23 The handlebody surface \( {H}_{g} \) has Euler characteristic \( \chi \left( {H}_{g}\right) = 2 - \) \( {2g} \), for all \( g \geq 0 \) . The cross-cap surface \( {C}_{g} \) has Euler characteristic \( \chi \left( {C}_{g}\right) = 2 - g \) , for all \( g \geq 1 \) .
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Proof. We have already seen that the Euler characteristic of the sphere is 2, so the result holds for \( {H}_{0} \) . For \( g \geq 1 \) consider the standard polygonal representation of \( {H}_{g} \) as a \( {4g} \) -gon with boundary label \( \left( {{a}_{1}{b}_{1}{a}_{1}^{-1}{b}_{1}^{-1}}\right) \cdots \left( {{a}_{g}{b}_{g}{a}_{g}^{-1}{b}_{g}^{-1}}\right) \) . One checks that all the corners come together at a single point, so our cell division of \( {H}_{g} \) has a single vertex. For instance, consider the 2-holed torus in Figure 7.5.24. Starting at the lower right-hand corner labeled (1), begin traversing the corner point in a clockwise direction. After hitting the \( {b}_{1} \) edge near its initial point, one reappears on the other \( {b}_{1} \) edge near its initial point. Keep circling the corner (according to the sequence indicated) until you return to the starting point. Notice that all eight corners are traversed before returning to the starting point. So the cell division determined by the polygon will have a single vertex. Since the \( {4g} \) edges are identified in pairs, the cell division has \( {2g} \) edges, and there is one face, the interior of the polygon. Thus, \( \chi \left( {H}_{g}\right) = 1 - {2g} + 1 = 2 - {2g} \) .\n\n\n\nFigure 7.5.24 This cell division of \( {H}_{2} \) has a single vertex.\n\nThe cross-cap surface \( {C}_{g} \) can be represented by the polygonal surface obtained by identifying the edges of a \( {2g} \) -gon according to the boundary label \( \left( {{a}_{1}{a}_{1}}\right) \cdots \left( {{a}_{g}{a}_{g}}\right) \) . Once again, all the corners come together at a single point in the edge identification, and the number of edges in the cell division is half the number of edges in the \( {2g} \) -gon. So the cell division determined by the polygon has one vertex, \( g \) edges, and one face. Thus, \( \chi \left( {C}_{g}\right) = 2 - g \) .
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Yes
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Theorem 7.6.4 Each surface admits one homogeneous, isotropic, and metric geometry. In particular, the sphere \( \left( {H}_{0}\right) \) and projective plane \( \left( {C}_{1}\right) \) admit elliptic geometry. The torus \( \left( {H}_{1}\right) \) and Klein bottle \( \left( {C}_{2}\right) \) admit Euclidean geometry. All handlebody surfaces \( {H}_{g} \) with \( g \geq 2 \) and all cross-cap surfaces \( {C}_{g} \) with \( g \geq 3 \) admit hyperbolic geometry.
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The sphere and projective plane admit elliptic geometry by construction: The space in elliptic geometry \( {is} \) the projective plane, and via stereographic projection, this is the geometry on \( {\mathbb{S}}^{2} \). The torus and Klein bottle are built from regular 4-gons (squares) whose edges are identified in such a way that all 4 corners come together at a point. In each case, if we place the square in the Euclidean plane all corner angles are \( \pi /2 \), so the sum of the angles is \( {2\pi } \), and our surfaces admit Euclidean geometry. Each handlebody surfaces \( {H}_{g} \) for \( g \geq 2 \) and each cross-cap surfaces \( {C}_{g} \) for \( g \geq 3 \) can be built from a regular \( n \) -gon where \( n \geq 6 \). Again, all \( n \) corners come together at a single point. A regular \( n \) -gon in the Euclidean plane has interior angle \( \left( {n - 2}\right) \pi /n \) radians, so the corner angles sum to \( \left( {n - 2}\right) \pi \) radians. This angle sum exceeds \( {2\pi } \) radians since \( n \geq 6 \). Placing a small version of this \( n \) -gon in the hyperbolic plane, the corner angle sum will be very nearly equal to \( \left( {n - 2}\right) \pi \) radians and will exceed \( {2\pi } \) radians, but as we expand the \( n \) -gon the corners approach ideal points and the corner angles sum will approach 0 radians. Thus, at some point the angle sum will equal \( {2\pi } \) radians on the nose, and the polygonal surface built from this precise \( n \) -gon admits hyperbolic geometry. Exercise 7.6.4 works through how to construct this precise \( n \) -gon.
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Yes
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If we make three slices in the two-holed torus we obtain two pairs of pants, as indicated in the following figure.
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We label our cuts so that we can stitch up our surface later. Match the \( {c}_{i},{d}_{i} \), and \( {e}_{i} \) edges to recover the two-holed torus. Any pair of pants can be cut into two hexagons by cutting along the three vertical seams in the pants. It follows that the two-holed torus can be constructed from four hexagons, with edges identified in pairs as indicated below. The four hexagons represent a cell division of \( {\mathrm{H}}_{2} \) having 6 vertices, 12 edges, and 4 faces. In the edge identification, corners come together in groups of 4, so we need each corner angle to equal \( {90}^{ \circ } \) in order to endow it with a homogeneous geometry. We know we can do this in the hyperbolic plane. Moreover, according to Theorem 5.4.19, there is freedom in choosing the dimensions of the hexagons. That is, for each triple of real numbers \( \left( {a, b, c}\right) \) there exists a right-angled hexagon in \( \mathbb{D} \) with alternate lengths \( a, b \), and \( c \) . So, there exists a two-holed torus for each combination of six seam lengths \( \left( {{a}_{1},{a}_{2},{a}_{3},{b}_{1},{b}_{2}\text{, and }{b}_{3}}\right) \) .
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Yes
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Theorem 7.6.7 Gauss-Bonnet. The area of a surface with constant curvature \( k \) and Euler characteristic \( \chi \) is given by the formula \( {kA} = {2\pi \chi } \) .
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Proof. The sphere with constant curvature \( k \) has radius equal to \( 1/\sqrt{k} \), and area equal to \( {4\pi }/k \) . Since the sphere has Euler characteristic 2, the Gauss-Bonnet formula holds in this case. The projective plane \( {\mathbb{P}}^{2} \) with curvature \( k \) has area equal to \( {2\pi }/k \), and Euler characteristic equal to 1, so the result holds in this case as well. The torus and the Klein bottle each have \( k = 0 \) and \( \chi = 0 \), so in this case the Gauss-Bonnet formula reduces to the true statement that \( 0 = 0 \) . Any surface of constant negative curvature cam be represented by a regular \( n \) -sided polygon with \( n \geq 6 \) . Furthermore, this polygon can be placed in the hyperbolic plane with curvature \( k < 0 \), so that its interior angles sum to \( {2\pi } \) radians. According to Theorem 7.4.4, \[ {kA} = {2\pi } - \left( {n - 2}\right) \pi = {2\pi }\left( {2 - \frac{n}{2}}\right) , \] where \( A \) is the area of the \( n \) -gon. Now, for \( g \geq 2,{H}_{g} \) is represented by a \( {4g} \) -gon so that \[ {kA} = {2\pi }\left( {2 - {2g}}\right) = {2\pi \chi }\left( {H}_{g}\right) . \] For \( g \geq 3,{C}_{g} \) is represented by a \( {2g} \) -gon so that \[ {kA} = {2\pi }\left( {2 - g}\right) = {2\pi \chi }\left( {C}_{g}\right) . \] This completes the proof.
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Yes
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Lemma 7.7.4 Suppose \( \sim \) is an equivalence relation on \( A \), and \( a \) and \( b \) are any two elements of \( A \) . Then either \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) have no elements in common, or they are equal sets.
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Proof. Suppose there is some element \( c \) that is in both \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) . We show \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) by arguing that each set is a subset of the other.\n\nThat \( \left\lbrack a\right\rbrack \) is a subset of \( \left\lbrack b\right\rbrack \) : Suppose \( x \) is in \( \left\lbrack a\right\rbrack \) . We must show that \( x \) is in \( \left\lbrack b\right\rbrack \) as well. Since \( x \) is in \( \left\lbrack a\right\rbrack, x \sim a \) . Since \( c \) is in \( \left\lbrack a\right\rbrack \) and in \( \left\lbrack b\right\rbrack, c \sim a \) and \( c \sim b \) . We may use these facts, along with transitivity and symmetry of the relation, to see that \( x \sim a \sim c \sim b \) . That is, \( x \) is in \( \left\lbrack b\right\rbrack \) . Therefore, everything in \( \left\lbrack a\right\rbrack \) is also in \( \left\lbrack b\right\rbrack \) .\n\nWe may repeat the argument above to show that \( \left\lbrack b\right\rbrack \) is a subset \( \left\lbrack a\right\rbrack \) . Thus, if \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) have any element in common, then they are entirely equal sets, and this completes the proof.
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Yes
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Example 7.7.8 \( {H}_{1} \) as quotient of \( \mathbb{C} \) .\n\nSuppose \( a \) and \( b \) are positive real numbers. Let \( \left\langle {{T}_{a},{T}_{bi}}\right\rangle \) be the group of homeomorphisms generated by the horizontal translation \( {T}_{a}\left( z\right) = z + a \) and the vertical translation \( {T}_{bi}\left( z\right) = z + {bi} \) .\n\nThis group contains all possible compositions of these two transformations and their inverses. Thus, the orbit of a point \( z \) consists of all complex numbers to which \( z \) can be sent by moving \( z \) horizontally by some integer multiple of \( a \) units, and vertically by some integer multiple of \( b \) units. An arbitrary transformation in \( \Gamma = \left\langle {{T}_{a},{T}_{bi}}\right\rangle \) has the form\n\n\[ T\left( z\right) = z + \left( {{ma} + {nbi}}\right) \]\n\nwhere \( m \) and \( n \) are integers. A fundamental domain for the orbit space consists of the rectangle with corners \( 0, a, a + {bi},{bi} \) . The resulting quotient space is homeomorphic to the torus. Notice that points on the boundary of this rectangle are identified in pairs. In fact, the fundamental domain, with\n\nits boundary point redundancies, corresponds precisely to our polygonal surface representation of the torus.
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If the space \( M \) has a metric and our group of homeomorphisms is sufficiently nice, then the resulting orbit space inherits a metric from the universal covering space \( M \) . To be sufficiently nice, we first need our homeomorphisms to be isometries. The group of isometries must also be fixed-point free and properly discontinuous. The group \( G \) is fixed-point free if each isometry in \( G \) (other than the identity map) has no fixed points. The group \( G \) is properly discontinuous if every \( x \) in \( X \) has an open 2-ball \( {U}_{x} \) about it whose images under all isometries in \( G \) are pairwise disjoint. The interested reader is encouraged to see [10] or [9] for more detail. If our group \( G \) is a fixed-point free, properly discontinuous group of isometries, then the resulting orbit space inherits a metric from \( M \) .\n\nConsider the quotient space in Example 7.7.7. The group here is a group of isometries, since rotations preserve Euclidean distance, but it is not fixed-point free. All maps in \( G \) have fixed points (rotation about the origin fixes 0 ). This prevents the quotient space from inheriting the geometry of its mother space. Indeed, a circle centered at \( \left\lbrack 0\right\rbrack \) with radius \( r \) would have circumference \( \frac{2\pi r}{4} \), which doesn't correspond to Euclidean geometry.\n\nThe group of isometries in the torus example is fixed-point free and properly discontinuous, so the following formula for the distance between two points \( \left\lbrack u\right\rbrack \) and \( \left\lbrack v\right\rbrack \) in the orbit space \( \mathbb{C}/\left\langle {{T}_{a},{T}_{bi}}\right\rangle \) is well-defined:\n\n\[ d\left( {\left\lbrack u\right\rbrack ,\left\lbrack v\right\rbrack }\right) = \min \{ \left| {z - w}\right| \mid z \in \left\lbrack u\right\rbrack, w \in \left\lbrack v\right\rbrack \} . \]\n\nFigure 7.7.9 depicts two points in the shaded fundamental domain, \( \left\lbrack u\right\rbrack \) and \( \left\lbrack v\right\rbrack \) . The distance between them equals the Euclidean distance in \( \mathbb{C} \) of the shortest path between any point in equivalence class \( \left\lbrack u\right\rbrack \) and any point in equivalence class \( \left\lbrack v\right\rbrack \) . There are many such nearest pairs, and one such pair is labeled in Figure 7.7.9 where \( z \) is in \( \left\lbrack u\right\rbrack \) and \( w \) is in \( \left\lbrack v\right\rbrack \) . Also drawn in the figure is a solid line (in two parts) that corresponds to the shortest path one would take within the fundamental domain to proceed from \( \left
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Yes
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Example 7.7.10 \( {\mathbb{P}}^{2} \) as quotient of \( {\mathbb{S}}^{2} \).
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Let \( {T}_{a} : {\mathbb{S}}^{2} \rightarrow {\mathbb{S}}^{2} \) be the antipodal map \( {T}_{a}\left( P\right) = - P \). This map is an isometry that sends each point on \( {\mathbb{S}}^{2} \) to the point diametrically opposed to it, so it is fixed-point free. Since \( {T}_{a}^{-1} = {T}_{a} \), the group generated by this map consists of just 2 elements: \( {T}_{a} \) and the identity map. The quotient space \( {\mathbb{S}}^{2}/\left\langle {T}_{a}\right\rangle \) is the projective plane.
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Yes
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Example 7.7.11 \( {\mathrm{H}}_{2} \) as quotient of \( {\mathbb{D}}^{2} \) .
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We may build a regular octagon in the hyperbolic plane whose interior angles equal \( \pi /4 \) radians. We may also find a hyperbolic transformation that takes an edge of this octagon to another edge. Labelling the edges as in the following diagram, let \( {T}_{a} \) be the hyperbolic isometry taking one \( a \) edge to the other, being careful to respect the edge orientations. We construct such a map by composing two hyperbolic reflections about hyperbolic lines: the hyperbolic line containing the first \( a \) edge, and the hyperbolic line \( m \) that bisects the \( b \) edge between the \( a \) edges. Since the two lines of reflection do not intersect, the resulting map in \( \mathcal{H} \) is a translation in \( \mathcal{H} \) and has no fixed points in \( \mathbb{D} \) (the fixed points are ideal points).\n\nDefine \( {T}_{b},{T}_{c} \), and \( {T}_{d} \) similarly and consider the group of isometries of \( \mathbb{D} \) generated by these four maps. This group is a fixed-point free, properly discontinuous group of isometries of \( \mathbb{D} \), so the resulting quotient space inherits hyperbolic geometry.\n\nThe distance between two points \( \left\lbrack u\right\rbrack \) and \( \left\lbrack v\right\rbrack \) in the quotient space is given by\n\n\[{d}_{H}\left( {\left\lbrack u\right\rbrack ,\left\lbrack v\right\rbrack }\right) = \min \left\{ {{d}_{H}\left( {z, w}\right) \mid z \in \left\lbrack u\right\rbrack, w \in \left\lbrack v\right\rbrack }\right\} .\n\]\n\nGeodesics in the quotient space are determined by geodesics in the hyperbolic plane \( \mathbb{D} \) .\n\nTopologically, the quotient space is homeomorphic to \( {H}_{2} \), and the octagon pictured above serves as a fundamental domain of the quotient space. Moving copies of this octagon by isometries in the group produces a tiling of \( \mathbb{D} \) by this octagon. Each copy of the octagon would serve equally well as a fundamental domain for the quotient space.
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Yes
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Show that the Dirichlet domain at any point of the torus in Example 7.7.8 is an \( a \) by \( b \) rectangle by completing the following parts.
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a. Construct an \( a \) by \( b \) rectangle to be the fundamental domain, and place eight copies of this rectangle around the fundamental domain as in Figure 7.7.9. Then plot a point \( x \) in the fundamental domain, and plot its image in each of the copies.\nb. For each image \( {x}^{\prime } \) of \( x \), construct the perpendicular bisector of the segment \( x{x}^{\prime } \) . The eight perpendicular bisectors enclose the Dirichlet domain based at \( x \) . Prove that the Dirichlet domain is also an \( a \) by \( b \) rectangle.
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No
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Proposition 1.3.1. For any \( a, b \in \mathbb{Q},\left| {a + b}\right| \leq \left| a\right| + \left| b\right| \) .
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Proof. If \( a + b \geq 0 \), then\n\n\[ \left| a\right| + \left| b\right| - \left| {a + b}\right| = \left| a\right| + \left| b\right| - a - b = \left( {\left| a\right| - a}\right) + \left( {\left| b\right| - b}\right) .\](1.3.15)\n\nBoth of the terms on the right are nonnegative by Exercise 1.3.8. Hence the sum is nonnegative and the proposition follows. If \( a + b < 0 \), then\n\n\[ \left| a\right| + \left| b\right| - \left| {a + b}\right| = \left| a\right| + \left| b\right| + a + b = \left( {\left| a\right| + a}\right) + \left( {\left| b\right| + b}\right) .\](1.3.16)\n\nAgain, both of the terms on the right are nonnegative by Exercise 1.3.8. Hence the sum is nonnegative and the theorem follows.\n\nQ.E.D.
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No
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Consider the set\n\n\[ A = \left\{ {a : a \in {\mathbb{Q}}^{ + },{a}^{2} < 2}\right\} . \]\n\nNow suppose \( s \in {\mathbb{Q}}^{ + } \) is the supremum of \( A \) . We must have either \( {s}^{2} < 2 \) , \( {s}^{2} > 2 \), or \( {s}^{2} = 2 \) .
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Suppose \( {s}^{2} < 2 \) and let \( \epsilon = 2 - {s}^{2} \) . By the archimedean property of \( \mathbb{Q} \), we may choose \( n \in {\mathbb{Z}}^{ + } \) such that\n\n\[ \frac{{2s} + 1}{n} < \epsilon \]\n\nfrom which it follows that\n\n\[ \frac{2s}{n} + \frac{1}{{n}^{2}} = \frac{{2s} + \frac{1}{n}}{n} \leq \frac{{2s} + 1}{n} < \epsilon . \]\n\nHence\n\n\[ {\left( s + \frac{1}{n}\right) }^{2} = {s}^{2} + \frac{2s}{n} + \frac{1}{{n}^{2}} < {s}^{2} + \epsilon = 2, \]\n\nwhich implies that \( s + \frac{1}{n} \in A \) . Since \( s < s + \frac{1}{n} \), this contradicts the assumption that \( s \) is an upper bound for \( A \) .\n\nSo now suppose \( {s}^{2} > 2 \) . Again let \( n \in {\mathbb{Z}}^{ + } \) and note that\n\n\[ {\left( s - \frac{1}{n}\right) }^{2} = {s}^{2} - \frac{2s}{n} + \frac{1}{{n}^{2}} \]\n\nIf we let \( \epsilon = {s}^{2} - 2 \), then we may choose \( n \in {\mathbb{Z}}^{ + } \) so that\n\n\[ \frac{2s}{n} < \epsilon \]\n\nIt follows that\n\n\[ {\left( s - \frac{1}{n}\right) }^{2} > {s}^{2} - \epsilon + \frac{1}{{n}^{2}} = 2 + \frac{1}{{n}^{2}} > 2. \]\n\nThus \( s - \frac{1}{n} \) is an upper bound for \( A \) and \( s - \frac{1}{n} < s \), contradicting the assumption that \( s = \sup A \) .\n\nThus we must have \( {s}^{2} = 2 \) . However, this is impossible in light of the following proposition. Hence we must conclude that \( A \) does not have a supremum.
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Yes
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Proposition 1.3.2. There does not exist a rational number \( s \) with the property that \( {s}^{2} = 2 \) .
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Proof. Suppose there exists \( s \in \mathbb{Q} \) such that \( {s}^{2} = 2 \) . Choose \( a, b \in {\mathbb{Z}}^{ + } \) so that \( a \) and \( b \) are relatively prime (that is, they have no factor other than 1 in common) and \( s = \frac{a}{b} \) . Then\n\n\[ \frac{{a}^{2}}{{b}^{2}} = 2 \]\n\nso \( {a}^{2} = 2{b}^{2} \) . Thus \( {a}^{2} \), and hence \( a \), is an even integer. So there exists \( c \in {\mathbb{Z}}^{ + } \) such that \( a = {2c} \) . Hence\n\n\[ 2{b}^{2} = {a}^{2} = 4{c}^{2} \]\n\nfrom which it follows that \( {b}^{2} = {2c} \), and so \( b \) is also an even integer. But this contradicts the assumption that \( a \) and \( b \) are relatively prime.\n\nQ.E.D.
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Yes
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We have\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\frac{1}{i} = 0 \]
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since, for any rational number \( \epsilon > 0 \) ,\n\n\[ \left| {\frac{1}{i} - 0}\right| = \frac{1}{i} < \epsilon \] \n\nfor any \( i > N \), where \( N \) is any integer larger than \( \frac{1}{\epsilon } \) .
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Yes
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Proposition 1.3.3. If \( {\left\{ {a}_{i}\right\} }_{i \in I} \) converges, then \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is a Cauchy sequence.
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Proof. Suppose \( \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} = L \) . Given \( \epsilon \in {\mathbb{Q}}^{ + } \), choose an integer \( N \) such that\n\n\[ \left| {{a}_{i} - L}\right| < \frac{\epsilon }{2} \]\n\n\( \left( {1.3.23}\right) \)\n\nfor all \( i > N \) . Then for any \( i, k > N \), we have\n\n\[ \left| {{a}_{i} - {a}_{k}}\right| = \left| {\left( {{a}_{i} - L}\right) + \left( {L - {a}_{k}}\right) }\right| \leq \left| {{a}_{i} - L}\right| + \left| {{a}_{k} - L}\right| < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon . \]\n\n\( \left( {1.3.24}\right) \)\n\nHence \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is a Cauchy sequence.\n\nQ.E.D.
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Yes
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Proposition 1.4.1. For any \( a, b \in \mathbb{R},\left| {a + b}\right| \leq \left| a\right| + \left| b\right| \) .
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Proof. If \( a + b \geq 0 \), then\n\n\[ \left| a\right| + \left| b\right| - \left| {a + b}\right| = \left| a\right| + \left| b\right| - a - b = \left( {\left| a\right| - a}\right) + \left( {\left| b\right| - b}\right) .\](1.4.25)\n\nBoth of the terms on the right are nonnegative by Exercise 1.4.10. Hence the sum is nonnegative and the proposition follows. If \( a + b < 0 \), then\n\n\[ \left| a\right| + \left| b\right| - \left| {a + b}\right| = \left| a\right| + \left| b\right| + a + b = \left( {\left| a\right| + a}\right) + \left( {\left| b\right| + b}\right) .\](1.4.26)\n\nAgain, both of the terms on the right are nonnegative by Exercise 1.4.10. Hence the sum is nonnegative and the proposition follows. Q.E.D.
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No
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Proposition 1.4.2. Given \( a \in {\mathbb{R}}^{ + } \), there exist \( r, s \in \mathbb{Q} \) such that \( 0 < r < a < s \) .
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Proof. Let \( \{ u{\} }_{i \in I} \) be a Cauchy sequence in the equivalence class of \( a \) . Since \( a > 0 \), there exists a rational \( \epsilon > 0 \) and an integer \( N \) such that \( {u}_{i} > \epsilon \) for all \( i > N \) . Let \( r = \frac{\epsilon }{2} \) . Then \( {u}_{i} - r > \frac{\epsilon }{2} \) for every \( i > N \), so \( a - r > 0 \), that is, \( 0 < r < a \) .\n\nNow choose an integer \( M \) so that \( \left| {{u}_{i} - {u}_{j}}\right| < 1 \) for all \( i, j > M \) . Let \( s = {u}_{M + 1} + 2 \) . Then\n\n\[ s - {u}_{i} = {u}_{M + 1} + 2 - {u}_{i} > 1 \]\n\nfor all \( i > M \) . Hence \( s > a \) .\n\nQ.E.D.
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Yes
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Proposition 1.4.3. \( \mathbb{R} \) is an archimedean ordered field.
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Proof. Given real numbers \( a \) and \( b \) with \( 0 < a < b \), let \( r \) and \( s \) be rational numbers for which \( 0 < r < a < b < s \) . Since \( \mathbb{Q} \) is a an archimedean field, there exists an integer \( n \) such that \( {nr} > s \) . Hence\n\n\[ \n{na} > {nr} > s > b\text{.}\n\]\n\n\( \left( {1.4.29}\right) \)\n\nQ.E.D.
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Yes
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Proposition 1.4.4. Given \( a, b \in \mathbb{R} \) with \( a < b \), there exists \( r \in \mathbb{Q} \) such that \( a < r < b \) .
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Proof. Let \( \{ u{\} }_{i \in I} \) be a Cauchy sequence in the equivalence class of \( a \) and let \( \{ v{\} }_{j \in J} \) be in the equivalence class of \( b \) . Since \( b - a > 0 \), there exists a rational \( \epsilon > 0 \) and an integer \( N \) such that \( {v}_{i} - {u}_{i} > \epsilon \) for all \( i > N \) . Now choose an integer \( M \) so that \( \left| {{u}_{i} - {u}_{j}}\right| < \frac{\epsilon }{4} \) for all \( i, j > M \) . Let \( r = {u}_{M + 1} + \frac{\epsilon }{2} \) . Then\n\n\[ r - {u}_{i} = {u}_{M + 1} + \frac{\epsilon }{2} - {u}_{i} \]\n\n\[ = \frac{\epsilon }{2} - \left( {{u}_{i} - {u}_{M + 1}}\right) \]\n\n\[ > \frac{\epsilon }{2} - \frac{\epsilon }{4} \]\n\n\[ = \frac{\epsilon }{4} \]\n\n\( \left( {1.4.30}\right) \)\n\nfor all \( i > M \) and\n\n\[ {v}_{i} - r = {v}_{i} - {u}_{M + 1} - \frac{\epsilon }{2} \]\n\n\[ = \left( {{v}_{i} - {u}_{i}}\right) - \left( {{u}_{M + 1} - {u}_{i}}\right) - \frac{\epsilon }{2} \]\n\n\[ > \epsilon - \frac{\epsilon }{4} - \frac{\epsilon }{2} \]\n\n\[ = \frac{\epsilon }{4} \]\n\n(1.4.31)\n\nfor all \( i \) larger than the larger of \( N \) and \( M \) . Hence \( a < r < b \) .
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Yes
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Theorem 1.4.5. Suppose \( A \subset \mathbb{R}, A \neq \varnothing \), has an upper bound. Then \( \sup A \) exists.
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Proof. Let \( a \in A \) and let \( b \) be an upper bound for \( A \) . Define sequences \( {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty } \) and \( {\left\{ {b}_{i}\right\} }_{i = 1}^{\infty } \) as follows: Let \( {a}_{1} = a \) and \( {b}_{1} = b \) . For \( i > 1 \), let\n\n\[ c = \frac{{a}_{i - 1} + {b}_{i - 1}}{2}. \]\n\nIf \( c \) is an upper bound for \( A \), let \( {a}_{i} = {a}_{i - 1} \) and let \( {b}_{i} = c \) ; otherwise, let \( {a}_{i} = c \) and \( {b}_{i} = {b}_{i - 1} \) . Then\n\n\[ \left| {{b}_{i} - {a}_{i}}\right| = \frac{\left| b - a\right| }{{2}^{i - 1}} \]\n\nfor \( i = 1,2,3,\ldots \) . Now, for \( i = 1,2,3,\ldots \), let \( {r}_{i} \) be a rational number such that \( {a}_{i} < {r}_{i} < {b}_{i} \) . Given any \( \epsilon > 0 \), we may choose \( N \) so that\n\n\[ {2}^{N} > \frac{\left| b - a\right| }{\epsilon } \]\n\nThen, whenever \( i > N \) and \( j > N \) ,\n\n\[ \left| {{r}_{i} - {r}_{j}}\right| < \left| {{b}_{N + 1} - {a}_{N + 1}}\right| = \frac{\left| b - a\right| }{{2}^{N}} < \epsilon . \]\n\nHence \( {\left\{ {r}_{i}\right\} }_{i = 1}^{\infty } \) is a Cauchy sequence. Let \( s \in \mathbb{R} \) be the equivalence class of \( {\left\{ {r}_{i}\right\} }_{i = 1}^{\infty } \) . Note that, for \( i = 1,2,3,\ldots ,{a}_{i} \leq s \leq {b}_{i} \) .\n\nNow if \( s \) is not an upper bound for \( A \), then there exists \( a \in A \) with \( a > s \) . Let \( \delta = a - s \) and choose an integer \( N \) such that\n\n\[ {2}^{N} > \frac{\left| b - a\right| }{\delta } \]\n\nThen\n\n\[ {b}_{N + 1} \leq s + \frac{\left| b - a\right| }{{2}^{N}} < s + \delta = a. \]\n\nBut, by construction, \( {b}_{N + 1} \) is an upper bound for \( A \) . Thus \( s \) must be an upper bound for \( A \) .\n\nNow suppose \( t \) is another upper bound for \( A \) and \( t < s \) . Let \( \delta = s - t \) and choose an integer \( N \) such that\n\n\[ {2}^{N} > \frac{\left| b - a\right| }{\delta } \]\n\nThen\n\n\[ {a}_{N + 1} \geq s - \frac{\left| b - a\right| }{{2}^{N}} > s - \delta = t, \]\n\nwhich implies that \( {a}_{N + 1} \) is an upper bound for \( A \) . But, by construction, \( {a}_{N + 1} \) is not an upper bound for \( A \) . Hence \( s \) must be the least upper bound for \( A \) , that is, \( s = \sup A \) . Q.E.D.
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Yes
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Theorem 2.1.1. If \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is a nondecreasing, bounded sequence of real numbers, then \( {\left\{ {a}_{i}\right\} }_{i \in I} \) converges.
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Proof. Since \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is bounded, the set of \( A = \left\{ {{a}_{i} : i \in I}\right\} \) has a supremum. Let \( L = \sup A \) . For any \( \epsilon > 0 \), there must exist \( N \in I \) such that \( {a}_{N} > L - \epsilon \) (or else \( L - \epsilon \) would be an upper bound for \( A \) which is smaller than \( L \) ). But then\n\n\[ L - \epsilon < {a}_{N} \leq {a}_{i} \leq L < L + \epsilon \]\n\nfor all \( i \geq N \), that is,\n\n\[ \left| {{a}_{i} - L}\right| < \epsilon \]\n\nfor all \( i \geq N \) . Thus \( {\left\{ {a}_{i}\right\} }_{i \in I} \) converges and\n\n\[ L = \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} \]
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Yes
|
Proposition 2.1.2. Suppose \( {\left\{ {a}_{i}\right\} }_{i \in I},{\left\{ {b}_{j}\right\} }_{j \in J} \), and \( {\left\{ {c}_{k}\right\} }_{k \in K} \) are sequences of real numbers for which there exists an integer \( N \) such that \( {a}_{i} \leq {c}_{i} \leq {b}_{i} \) whenever \( i > N \) . If\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{b}_{i} \]\n\n\( \left( {2.1.16}\right) \)\n\nthen\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{c}_{i} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{b}_{i} \]\n\n\( \left( {2.1.17}\right) \)
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Proof. Let \( L = \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{b}_{i} \) . Suppose \( L \) is finite. Given \( \epsilon > 0 \), there exists an integer \( M \) such that\n\n\[ \left| {{a}_{i} - L}\right| < \frac{\epsilon }{3} \]\n\n(2.1.18)\n\nand\n\n\[ \left| {{b}_{i} - L}\right| < \frac{\epsilon }{3} \]\n\n\( \left( {2.1.19}\right) \)\n\nwhenever \( i > M \) . Then\n\n\[ \left| {{a}_{i} - {b}_{i}}\right| \leq \left| {{a}_{i} - L}\right| + \left| {L - {b}_{i}}\right| < \frac{\epsilon }{3} + \frac{\epsilon }{3} = \frac{2\epsilon }{3} \]\n\n\( \left( {2.1.20}\right) \)\n\nwhenever \( i > M \) . Let \( K \) be the larger of \( N \) and \( M \) . Then\n\n\[ \left| {{c}_{i} - L}\right| \leq \left| {{c}_{i} - {b}_{i}}\right| + \left| {{b}_{i} - L}\right| \leq \left| {{a}_{i} - {b}_{i}}\right| + \left| {{b}_{i} - L}\right| < \frac{2\epsilon }{3} + \frac{\epsilon }{3} = \epsilon \]\n\n\( \left( {2.1.21}\right) \)\n\nwhenever \( i > K \) . Thus \( \mathop{\lim }\limits_{{i \rightarrow \infty }}{c}_{i} = L \) . The result when \( L \) is infinite is a consequence of the next two exercises.\n\nQ.E.D.
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No
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Proposition 2.1.3. Suppose \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is a sequence for which\n\n\[ \mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} \]\n\n\( \left( {2.1.24}\right) \)\n\nThen\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} \]\n\n\( \left( {2.1.25}\right) \)
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Proof. Let \( {u}_{i} = \sup \left\{ {{a}_{k} : k \geq i}\right\} \) and \( {l}_{i} = \inf \left\{ {{a}_{k} : k \geq i}\right\} \) . Then \( {l}_{i} \leq {a}_{i} \leq {u}_{i} \) for all \( i \in I \) . Now\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{l}_{i} = \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\lim }\limits_{{i \rightarrow \infty }}{u}_{i}, \]\n\n(2.1.26)\n\nso the result follows from the squeeze theorem.\n\nQ.E.D.
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Yes
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Theorem 2.1.4. Suppose \( {\left\{ {a}_{i}\right\} }_{i \in I} \) is a Cauchy sequence in \( \mathbb{R} \) . Then \( {\left\{ {a}_{i}\right\} }_{i \in I} \) converges to a limit \( L \in \mathbb{R} \) .
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Proof. Let \( {u}_{i} = \sup \left\{ {{a}_{k} : k \geq i}\right\} \) and \( {l}_{i} = \inf \left\{ {{a}_{k} : k \geq i}\right\} \) . Given any \( \epsilon > 0 \) , there exists \( N \in \mathbb{Z} \) such that \( \left| {{a}_{i} - {a}_{j}}\right| < \epsilon \) for all \( i, j > N \) . Thus, for all \( i, j > N \) , \( {a}_{i} < {a}_{j} + \epsilon \), and so\n\n\[ \n{a}_{i} \leq \inf \left\{ {{a}_{j} + \epsilon : j \geq i}\right\} = {l}_{i} + \epsilon \n\]\n\n(2.1.28)\n\nfor all \( i > N \) . Since \( {\left\{ {l}_{i}\right\} }_{i \in I} \) is a nondecreasing sequence,\n\n\[ \n{a}_{i} \leq \sup \left\{ {{l}_{i} + \epsilon : i \in I}\right\} = \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} + \epsilon \n\]\n\n\( \left( {2.1.29}\right) \)\n\nfor all \( i > N \) . Hence\n\n\[ \n{u}_{i} = \sup \left\{ {{a}_{k} : k \geq i}\right\} \leq \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} + \epsilon \n\]\n\n\( \left( {2.1.30}\right) \)\n\nfor all \( i > N \) . Thus\n\n\[ \n\mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} = \inf \left\{ {{u}_{i} : i \in I}\right\} \leq \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} + \epsilon .\n\]\n\n\( \left( {2.1.31}\right) \)\n\nSince \( \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} \leq \mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} \), it follows that\n\n\[ \n\left| {\mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} - \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i}}\right| \leq \epsilon \n\]\n\n\( \left( {2.1.32}\right) \)\n\nSince this is true for every \( \epsilon > 0 \), we have \( \mathop{\limsup }\limits_{{i \rightarrow \infty }}{a}_{i} = \mathop{\liminf }\limits_{{i \rightarrow \infty }}{a}_{i} \), and so \( {\left\{ {a}_{i}\right\} }_{i \in I} \) converges by Proposition 2.1.3.\n\nQ.E.D.
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Yes
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Proposition 2.1.5. Suppose \( {\left\{ {x}_{i}\right\} }_{i \in I} \) is a convergent sequence in \( \mathbb{R},\alpha \) is a real number, and \( L = \mathop{\lim }\limits_{{i \rightarrow \infty }}{x}_{i} \) . Then the sequence \( {\left\{ \alpha {x}_{i}\right\} }_{i \in I} \) converges and\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\alpha {x}_{i} = {\alpha L} \]
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Proof. If \( \alpha = 0 \), then \( {\left\{ \alpha {x}_{i}\right\} }_{i \in I} \) clearly converges to 0 . So assume \( \alpha \neq 0 \) . Given \( \epsilon > 0 \), choose an integer \( N \) such that\n\n\[ \left| {{x}_{i} - L}\right| < \frac{\epsilon }{\left| \alpha \right| } \]\n\nwhenever \( i > N \) . Then for any \( i > N \) we have\n\n\[ \left| {\alpha {x}_{i} - {\alpha L}}\right| = \left| \alpha \right| \left| {{x}_{i} - L}\right| < \left| \alpha \right| \frac{\epsilon }{\left| \alpha \right| } = \epsilon . \]\n\nThus \( \mathop{\lim }\limits_{{i \rightarrow \infty }}\alpha {x}_{i} = {\alpha L} \) .
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Yes
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Proposition 2.1.8. Suppose \( {\left\{ {x}_{i}\right\} }_{i \in I} \) and \( {\left\{ {y}_{i}\right\} }_{i \in I} \) are convergent sequences in \( \mathbb{R} \) with \( L = \mathop{\lim }\limits_{{i \rightarrow \infty }}{x}_{i}, M = \mathop{\lim }\limits_{{i \rightarrow \infty }}{y}_{i} \), and \( {y}_{i} \neq 0 \) for all \( i \in I \) . If \( M \neq 0 \), then the sequence \( {\left\{ \frac{{x}_{i}}{{y}_{i}}\right\} }_{i \in I} \) converges and\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\frac{{x}_{i}}{{y}_{i}} = \frac{L}{M} \]
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Proof. Since \( M \neq 0 \) and \( M = \mathop{\lim }\limits_{{i \rightarrow \infty }}{y}_{i} \), we may choose an integer \( N \) such that\n\n\[ \left| {y}_{i}\right| > \frac{\left| M\right| }{2} \]\n\n(2.1.39)\n\nwhenever \( i > N \) . Let \( B \) be an upper bound for \( \left\{ {\left| {x}_{i}\right| : i \in I}\right\} \cup \left\{ {\left| {y}_{i}\right| : i \in I}\right\} \) . Given any \( \epsilon > 0 \), we may choose an integer \( P \) such that\n\n\[ \left| {{x}_{i} - L}\right| < \frac{{M}^{2}\epsilon }{4B} \]\n\n(2.1.40)\n\nand\n\n\[ \left| {{y}_{i} - M}\right| < \frac{{M}^{2}\epsilon }{4B} \]\n\n(2.1.41)\n\nwhenever \( i > P \) . Let \( K \) be the larger of \( N \) and \( P \) . Then, for any \( i > K \), we\n\nhave\n\n\[ \left| {\frac{{x}_{i}}{{y}_{i}} - \frac{L}{M}}\right| = \frac{\left| {x}_{i}M - {y}_{i}L\right| }{\left| {y}_{i}M\right| } \]\n\n\[ = \frac{\left| {x}_{i}M - {x}_{i}{y}_{i} + {x}_{i}{y}_{i} - {y}_{i}L\right| }{\left| {y}_{i}M\right| } \]\n\n\[ \leq \frac{\left| {x}_{i}\right| \left| {M - {y}_{i}}\right| + \left| {y}_{i}\right| \left| {{x}_{i} - L}\right| }{\left| {y}_{i}M\right| } \]\n\n\[ < \frac{B\frac{{M}^{2}\epsilon }{4B} + B\frac{{M}^{2}\epsilon }{4B}}{\frac{{M}^{2}}{2}} \]\n\n\[ = \epsilon \text{.} \]\n\n(2.1.42)\n\nThus\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\frac{{x}_{i}}{{y}_{i}} = \frac{L}{M} \]\n\n(2.1.43)\n\nQ.E.D.
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Yes
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Example 2.1.1. We may combine the properties of this section to compute\n\n\\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{5{n}^{3} + {3n} - 6}{2{n}^{3} + 2{n}^{2} - 7} = \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{5 + \\frac{3}{{n}^{2}} - \\frac{6}{{n}^{3}}}{2 + \\frac{2}{n} - \\frac{7}{{n}^{3}}} \n\\]\n
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\n\\[ \n= \\frac{\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}5 + 3\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{1}{{n}^{2}} - 6\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{1}{{n}^{3}}}{\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}2 + 2\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{1}{n} - 7\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{1}{{n}^{3}}} \n\\]\n\n\\[ \n= \\frac{5 + 0 + 0}{2 + 0 + 0} \n\\]\n\n\\[ \n= \\frac{5}{2}\\text{.} \n\\]\n
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Yes
|
Proposition 2.1.9. Suppose \( {\left\{ {x}_{i}\right\} }_{i \in I} \) is a convergent sequence of nonnegative real numbers with \( L = \mathop{\lim }\limits_{{i \rightarrow \infty }}{x}_{i} \) . Then the sequence \( {\left\{ \sqrt{{x}_{i}}\right\} }_{i \in I} \) converges and\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\sqrt{{x}_{i}} = \sqrt{L} \]
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Proof. Let \( \epsilon > 0 \) be given. Suppose \( L > 0 \) and note that\n\n\[ \left| {{x}_{i} - L}\right| = \left| {\sqrt{{x}_{i}} - \sqrt{L}}\right| \left| {\sqrt{{x}_{i}} + \sqrt{L}}\right| \]\n\nimplies that\n\n\[ \left| {\sqrt{{x}_{i}} - \sqrt{L}}\right| = \frac{\left| {x}_{i} - L\right| }{\left| \sqrt{{x}_{i}} + \sqrt{L}\right| } \]\n\nfor any \( i \in I \) . Choose an integer \( N \) such that\n\n\[ \left| {{x}_{i} - L}\right| < \sqrt{L}\epsilon \]\n\nwhenever \( i > N \) . Then, for any \( i > N \), \n\n\[ \left| {\sqrt{{x}_{i}} - \sqrt{L}}\right| = \frac{\left| {x}_{i} - L\right| }{\left| \sqrt{{x}_{i}} + \sqrt{L}\right| } < \frac{\sqrt{L}\epsilon }{\sqrt{L}} = \epsilon . \]\n\nHence \( \mathop{\lim }\limits_{{i \rightarrow \infty }}\sqrt{{x}_{i}} = \sqrt{L} \).\n\nIf \( L = 0,\mathop{\lim }\limits_{{i \rightarrow \infty }}{x}_{i} = 0 \), so we may choose an integer \( N \) such that \( \left| {x}_{i}\right| < {\epsilon }^{2} \) for all \( i > N \) . Then\n\n\[ \left| \sqrt{{x}_{i}}\right| < \epsilon \]\n\nwhenever \( i > N \), so \( \mathop{\lim }\limits_{{i \rightarrow \infty }}\sqrt{{x}_{i}} = 0 \).\n\nQ.E.D.
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Yes
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Proposition 2.1.10. If \( x \in \mathbb{R} \) and \( \left| x\right| < 1 \), then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n} = 0 \]
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Proof. We first assume \( x \geq 0 \) . Then the sequence \( {\left\{ {x}^{n}\right\} }_{n = 1}^{\infty } \) is nonincreasing and bounded below by 0 . Hence the sequence converges. Let \( L = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n} \) . Then\n\n\[ L = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n} = x\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n - 1} = {xL}, \]\n\nfrom which it follows that \( L\left( {1 - x}\right) = 0 \) . Since \( 1 - x > 0 \), we must have \( L = 0 \) . The result for \( x < 0 \) follows from the next exercise.\n\nQ.E.D.
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No
|
Proposition 2.1.11. Suppose \( \Lambda \) is the set of all subsequential limits of the sequence \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) . Then \( \Lambda \neq \varnothing \) .
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Proof. By the previous exercise, the proposition is true if \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) is not bounded. So suppose \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) is bounded and choose real numbers \( a \) and \( b \) such that \( a \leq {x}_{i} \leq b \) for all \( i \geq m \) . Construct sequences \( {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty } \) and \( {\left\{ {b}_{i}\right\} }_{i = 1}^{\infty } \) as follows: Let \( {a}_{1} = a \) and \( {b}_{1} = b \) . For \( i \geq 1 \), let\n\n\[ c = \frac{{a}_{i - 1} + {b}_{i - 1}}{2}. \]\n\nIf there exists an integer \( N \) such that \( {a}_{i - 1} \leq {x}_{j} \leq c \) for all \( j > N \), let \( {a}_{i} = {a}_{i - 1} \) and \( {b}_{i} = c \) ; otherwise, let \( {a}_{i} = c \) and \( {b}_{i} = {b}_{i - 1} \) . Let \( {n}_{1} = m \) and, for \( k = 2,3,4,\ldots \), let \( {n}_{k} \) be the smallest integer for which \( {n}_{k} > {n}_{k - 1} \) and \( {a}_{k} \leq {x}_{{n}_{k}} \leq {b}_{k} \) . Then \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) is a Cauchy sequence which is a subsequence of \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) . Thus \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) converges and \( \Lambda \neq \varnothing \) .\n\nQ.E.D.
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Yes
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Proposition 2.1.12. Let \( \Lambda \) be the set of subsequential limits of a sequence \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) . Then\n\n\[ \mathop{\limsup }\limits_{{i \rightarrow \infty }}{x}_{i} = \sup \Lambda \]
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Proof. Let \( s = \sup \Lambda \) and, for \( i \geq m,{u}_{i} = \sup \left\{ {{x}_{j} : j \geq i}\right\} \) . Now since \( {x}_{j} \leq {u}_{i} \) for all \( j \geq i \), it follows that \( \lambda \leq {u}_{i} \) for every \( \lambda \in \Lambda \) and \( i \geq m \) . Hence, from the previous exercise, \( s \leq \inf \left\{ {{u}_{i} : i \geq m}\right\} = \mathop{\limsup }\limits_{{i \rightarrow \infty }}{x}_{i} \) .\n\nNow suppose \( s < \mathop{\limsup }\limits_{{i \rightarrow \infty }}{x}_{i} \) . Then there exists a real number \( t \) such that \( s < t < \limsup {x}_{i} \) . In particular, \( t < {u}_{i} \) for every \( i \geq m \) . Let \( {n}_{1} \) be the smallest integer for which \( {n}_{1} \geq m \) and \( {x}_{{n}_{1}} > t \) . For \( k = 2,3,4,\ldots \), let \( {n}_{k} \) be the smallest integer for which \( {n}_{k} > {n}_{k - 1} \) and \( {x}_{{n}_{k}} > t \) . Then \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) is a subsequence of \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \) which has a subsequential limit \( \lambda \geq t \) . Since \( \lambda \) is also then a subsequential limit of \( {\left\{ {x}_{i}\right\} }_{i = m}^{\infty } \), we have \( \lambda \in \Lambda \) and \( \lambda \geq t > s \), contradicting \( s = \sup \Lambda \) . Hence we must have \( \mathop{\limsup }\limits_{{i \rightarrow \infty }}{x}_{i} = \sup \Lambda \) .\n\nQ.E.D.
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Yes
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Proposition 2.2.1. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) converges. Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) .
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Proof. Let \( {s}_{n} = \mathop{\sum }\limits_{{i = m}}^{n}{a}_{i} \) and \( s = \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} \) . Since \( {a}_{n} = {s}_{n} - {s}_{n - 1} \), we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n} - {s}_{n - 1}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} - \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n - 1} = s - s = 0. \]
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Yes
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Proposition 2.2.2. For any real number \( x \) with \( \left| x\right| < 1 \) , \n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n} = \frac{1}{1 - x} \]
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Proof. If \( {s}_{n} = \mathop{\sum }\limits_{{i = 0}}^{n}{x}^{i} \), then, by the previous exercise, \n\n\[ {s}_{n} = \frac{1 - {x}^{n + 1}}{1 - x} \] \n\nHence \n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1 - {x}^{n + 1}}{1 - x} = \frac{1}{1 - x}. \] \n\nQ.E.D.
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Yes
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Proposition 2.2.3. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) and \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) are infinite series for which there exists an integer \( N \) such that \( 0 \leq {a}_{i} \leq {b}_{i} \) whenever \( i \geq N \) . If \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) converges, then \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) converges.
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Proof. By Exercise 2.2.3 We need only show that \( \mathop{\sum }\limits_{{i = N}}^{\infty }{a}_{i} \) converges. Let \( {s}_{n} \) be the \( n \) th partial sum of \( \mathop{\sum }\limits_{{i = N}}^{\infty }{a}_{i} \) and let \( {t}_{n} \) be the \( n \) th partial sum of \( \mathop{\sum }\limits_{{i = N}}^{\infty }{b}_{i} \) . Now\n\n\[ \n{s}_{n + 1} - {s}_{n} = {a}_{n + 1} \geq 0 \n\]\n\n\( \left( {2.2.8}\right) \)\n\nfor every \( n \geq N \), so \( {\left\{ {s}_{n}\right\} }_{n = N}^{\infty } \) is a nondecreasing sequence. Moreover,\n\n\[ \n{s}_{n} \leq {t}_{n} \leq \mathop{\sum }\limits_{{i = N}}^{\infty }{b}_{i} < + \infty \n\]\n\n\( \left( {2.2.9}\right) \)\n\nfor every \( n \geq N \) . Thus \( {\left\{ {s}_{n}\right\} }_{n = N}^{\infty } \) is a nondecreasing, bounded sequence, and so converges. Q.E.D.
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No
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Proposition 2.2.4. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) and \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) are infinite series for which there exists an integer \( N \) such that \( 0 \leq {a}_{i} \leq {b}_{i} \) whenever \( i \geq N \) . If \( \mathop{\sum }\limits_{{i = k}}^{\infty }{a}_{i} \) diverges, then \( \mathop{\sum }\limits_{{i = m}}^{\infty }{b}_{i} \) diverges.
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Proof. By Exercise 2.2.3 we need only show that \( \mathop{\sum }\limits_{{i = N}}^{\infty }{b}_{i} \) diverges. Let \( {s}_{n} \) be the \( n \) th partial sum of \( \mathop{\sum }\limits_{{i = N}}^{\infty }{a}_{i} \) and let \( {t}_{n} \) be the \( n \) th partial sum of \( \mathop{\sum }\limits_{{i = N}}^{\infty }{b}_{i} \) . Now \( {\left\{ {s}_{n}\right\} }_{n = N}^{\infty } \) is a nondecreasing sequence which diverges, and so we must have \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = + \infty \) . Thus given any real number \( M \) there exists an integer \( L \) such that \[ M < {s}_{n} \leq {t}_{n} \] whenever \( n > L \) . Hence \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{t}_{n} = + \infty \) and \( \mathop{\sum }\limits_{{i = m}}^{\infty }{b}_{i} \) diverges. Q.E.D.
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No
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Consider the infinite series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{3!} + \frac{1}{4!} + \cdots . \]
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Now for \( n = 1,2,3,\ldots \), we have\n\n\[ 0 < \frac{1}{n!} \leq \frac{1}{{2}^{n - 1}} \]\n\nSince\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{2}^{n - 1}} \]\n\nconverges, it follows that\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} \]\n\nconverges. Moreover,\n\n\[ 2 < \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} < 1 + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{2}^{n - 1}} = 1 + \frac{1}{1 - \frac{1}{2}} = 3. \]\n\nWe let\n\n\[ e = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} \]
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Yes
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Proposition 2.2.5. \( e \notin \mathbb{Q} \) .
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Proof. Suppose \( e = \frac{p}{q} \) where \( p, q \in {\mathbb{Z}}^{ + } \) . Let\n\n\[ a = q!\left( {e - \mathop{\sum }\limits_{{i = 0}}^{q}\frac{1}{n!}}\right) .\n\]\n\n\( \left( {2.2.12}\right) \)\n\nThen \( a \in {\mathbb{Z}}^{ + } \) since \( q!e = \left( {q - 1}\right) !p \) and \( n! \) divides \( q! \) when \( n \leq q \) . At the same\n\ntime\n\n\[ a = q!\left( {\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} - \mathop{\sum }\limits_{{i = 0}}^{q}\frac{1}{n!}}\right)\n\]\n\n\[ = q!\mathop{\sum }\limits_{{n = q + 1}}^{\infty }\frac{1}{n!}\n\]\n\n\[ = \left( {\frac{1}{q + 1} + \frac{1}{\left( {q + 1}\right) \left( {q + 2}\right) } + \frac{1}{\left( {q + 1}\right) \left( {q + 2}\right) \left( {q + 3}\right) } + \cdots }\right)\n\]\n\n\[ = \frac{1}{q + 1}\left( {1 + \frac{1}{q + 2} + \frac{1}{\left( {q + 2}\right) \left( {q + 3}\right) } + \cdots }\right)\n\]\n\n\[ < \frac{1}{q + 1}\left( {1 + \frac{1}{q + 1} + \frac{1}{{\left( q + 1\right) }^{2}} + \cdots }\right)\n\]\n\n\[ = \frac{1}{q + 1}\mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{\left( q + 1\right) }^{n}}\n\]\n\n\[ = \frac{1}{q + 1}\left( \frac{1}{1 - \frac{1}{q + 1}}\right)\n\]\n\n\[ = \frac{1}{q}\text{.} \]\n\n\( \left( {2.2.13}\right) \)\n\nSince this is impossible, we conclude that no such integers \( p \) and \( q \) exist.\n\nQ.E.D.
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Yes
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Proposition 2.2.6. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) and \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) are infinite series for which there exists an integer \( N \) and a real number \( M > 0 \) such that \( 0 \leq {a}_{i} \leq M{b}_{i} \) whenever \( i \geq N \) . If \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) converges, then \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) converges.
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Proof. Since \( \mathop{\sum }\limits_{{i = k}}^{\infty }M{b}_{i} \) converges whenever \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) does, the result follows from the comparison test. Q.E.D.
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Yes
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Proposition 2.2.7. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) and \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) are infinite series for which there exists an integer \( N \) and a real number \( M > 0 \) such that \( 0 \leq {a}_{i} \leq M{b}_{i} \) whenever \( i \geq N \) . If \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) diverges, then \( \mathop{\sum }\limits_{{i = k}}^{\infty }{b}_{i} \) diverges.
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Proof. By the comparison test, \( \mathop{\sum }\limits_{{i = m}}^{\infty }M{b}_{i} \) diverges. Hence, by the previous exercise, \( \mathop{\sum }\limits_{{i = m}}^{\infty }{b}_{i} \) also diverges. Q.E.D.
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No
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Lemma 3.1.1. With the notation as above,\n\n\[ \n{s}_{n} \leq x < {s}_{n} + \frac{1}{{2}^{n}} \n\]\n\nfor \( n = 1,2,3,\ldots \)
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Proof. Since\n\n\[ \n{s}_{1} = \left\{ \begin{array}{ll} 0, & \text{ if }0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & \text{ if }\frac{1}{2} \leq x < 1 \end{array}\right.\n\]\n\nit is clear that \( {s}_{1} \leq x < {s}_{1} + \frac{1}{2} \) . So suppose \( n > 1 \) and \( {s}_{n - 1} \leq x < {s}_{n - 1} + \frac{1}{{2}^{n - 1}} \) . If \( {s}_{n - 1} + \frac{1}{{2}^{n}} \leq x \), then \( {a}_{n} = 1 \) and\n\n\[ \n{s}_{n} = {s}_{n - 1} + \frac{1}{{2}^{n}} \leq x < {s}_{n - 1} + \frac{1}{{2}^{n - 1}} = {s}_{n - 1} + \frac{1}{{2}^{n}} + \frac{1}{{2}^{n}} = {s}_{n} + \frac{1}{{2}^{n}}.\n\]\n\nIf \( x < {s}_{n - 1} + \frac{1}{{2}^{n}} \), then \( {a}_{n} = 0 \) and\n\n\[ \n{s}_{n} = {s}_{n - 1} \leq x < {s}_{n - 1} + \frac{1}{{2}^{n}} = {s}_{n} + \frac{1}{{2}^{n}}.\n\]\n\nQ.E.D.
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Yes
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Proposition 3.1.2. With the notation as above, \[ x = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{2}^{n}} \]
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Proof. Given \( \epsilon > 0 \), choose an integer \( N \) such that \( \frac{1}{{2}^{N}} < \epsilon \) . Then, for any \( n > N \), it follows from the lemma that \[ \left| {{s}_{n} - x}\right| < \frac{1}{{2}^{n}} < \frac{1}{{2}^{N}} < \epsilon \] \( \left( {3.1.12}\right) \) Hence \[ x = \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{2}^{n}}. \] (3.1.13) Q.E.D.
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Yes
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Lemma 3.1.3. With the notation as above, given any integer \( N \) there exists an integer \( n > N \) such that \( {a}_{n} = 0 \) .
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Proof. If \( {a}_{n} = 1 \) for \( n = 1,2,3,\ldots \), then\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{2}^{n}} = 1 \n\]\n\n(3.1.14)\n\ncontradicting the assumption that \( 0 \leq x < 1 \) . Now suppose there exists an integer \( N \) such that \( {a}_{N} = 0 \) but \( {a}_{n} = 1 \) for every \( n > N \) . Then\n\n\[ \nx = {s}_{N} + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{1}{{2}^{n}} = {s}_{N - 1} + \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{1}{{2}^{n}} = {s}_{N - 1} + \frac{1}{{2}^{N}}, \n\]\n\n\( \left( {3.1.15}\right) \)\n\nimplying that \( {a}_{N} = 1 \), and thus contradicting the assumption that \( {a}_{N} = 0 \) .\n\nQ.E.D.
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Yes
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Proposition 3.2.1. Suppose \( A \) and \( B \) are countable sets. Then the set \( C = \) \( A \cup B \) is countable.
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Proof. Suppose \( A \) and \( B \) are disjoint, that is, \( A \cap B = \varnothing \) . Let \( \varphi : {\mathbb{Z}}^{ + } \rightarrow A \) and \( \psi : {\mathbb{Z}}^{ + } \rightarrow B \) be one-to-one correspondences. Define \( \tau : {\mathbb{Z}}^{ + } \rightarrow C \) by\n\n\[ \tau \left( n\right) = \left\{ \begin{array}{ll} \varphi \left( \frac{n + 1}{2}\right) , & \text{ if }n\text{ is odd,} \\ \psi \left( \frac{n}{2}\right) , & \text{ if }n\text{ is even. } \end{array}\right. \]\n\nThen \( \tau \) is a one-to-one correspondence, showing that \( C \) is countable.\n\nIf \( A \) and \( B \) are not disjoint, then \( \tau \) is onto but not one-to-one. However, in that case \( C \) has the cardinality of an infinite subset of \( {\mathbb{Z}}^{ + } \), and so is countable.\n\nQ.E.D.
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Yes
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Proposition 3.2.2. Suppose \( A \) and \( B \) are countable. Then \( C = A \times B \) is countable.
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Proof. Let \( \varphi : {\mathbb{Z}}^{ + } \rightarrow A \) and \( \psi : {\mathbb{Z}}^{ + } \rightarrow B \) be one-to-one correspondences. Let \( {a}_{i} = \varphi \left( i\right) \) and \( {b}_{i} = \psi \left( i\right) \) . Define \( \tau : {\mathbb{Z}}^{ + } \rightarrow C \) by letting\n\n\[ \tau \left( 1\right) = \left( {{a}_{1},{b}_{1}}\right) \]\n\n\[ \tau \left( 2\right) = \left( {{a}_{1},{b}_{2}}\right) \]\n\n\[ \tau \left( 3\right) = \left( {{a}_{2},{b}_{1}}\right) \]\n\n\[ \tau \left( 4\right) = \left( {{a}_{1},{b}_{3}}\right) \]\n\n\[ \tau \left( 5\right) = \left( {{a}_{2},{b}_{2}}\right) \]\n\n\[ \tau \left( 6\right) = \left( {{a}_{3},{b}_{1}}\right) \]\n\n\[ \tau \left( 7\right) = \left( {{a}_{1},{b}_{4}}\right) \]\n\n\[ \vdots \; = \;\vdots \]\n\nThat is, form the infinite matrix with \( \left( {{a}_{i},{b}_{j}}\right) \) in the \( i \) th row and \( j \) th column, and then count the entries by reading down the diagonals from right to left. Then \( \tau \) is a one-to-one correspondence and \( C \) is countable.\n\nQ.E.D.
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Yes
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Proposition 3.2.3. \( \mathbb{Q} \) is countable.
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Proof. By the previous proposition, \( \mathbb{Z} \times \mathbb{Z} \) is countable. Let\n\n\[ A = \{ \left( {p, q}\right) : p, q \in \mathbb{Z}, q > 0, p\text{ and }q\text{ relatively prime }\} . \]\n\nThen \( A \) is infinite and \( A \subset \mathbb{Z} \times \mathbb{Z} \), so \( A \) is countable. But clearly \( \left| \mathbb{Q}\right| = \left| A\right| \), so \( \mathbb{Q} \) is countable. Q.E.D.
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Yes
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Proposition 3.2.4. Suppose for each \( i \in {\mathbb{Z}}^{ + },{A}_{i} \) is countable. Then\n\n\[ B = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} \]\n\nis countable.
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Proof. Suppose the sets \( {A}_{i}, i \in {\mathbb{Z}}^{ + } \), are pairwise disjoint, that is, \( {A}_{i} \cap {A}_{j} = \varnothing \) for all \( i, j \in {\mathbb{Z}}^{ + } \) . For each \( i \in {\mathbb{Z}}^{ + } \), let \( {\varphi }_{i} : {\mathbb{Z}}^{ + } \rightarrow {A}_{i} \) be a one-to-one correspondence. Then \( \psi : {\mathbb{Z}}^{ + } \times {\mathbb{Z}}^{ + } \rightarrow B \) defined by\n\n\[ \psi \left( {i, j}\right) = {\varphi }_{i}\left( j\right) \]\n\nis a one-to-one correspondence, and so \( \left| B\right| = \left| {{\mathbb{Z}}^{ + } \times {\mathbb{Z}}^{ + }}\right| = {\aleph }_{0} \) .\n\nIf the sets \( {A}_{i}, i \in {\mathbb{Z}}^{ + } \), are not disjoint, then \( \psi \) is onto but not one-to-one. But then there exists a subset \( P \) of \( {\mathbb{Z}}^{ + } \times {\mathbb{Z}}^{ + } \) such that \( \psi : P \rightarrow B \) is a one-to-one correspondence. Since \( P \) is an infinite subset of a countable set, \( P \) is countable and so \( \left| B\right| = {\aleph }_{0} \) .\n\nQ.E.D.
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Yes
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If \( A = \{ 1,2,3\} \), then
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\[ \mathcal{P}\left( A\right) = \{ \varnothing ,\{ 1\} ,\{ 2\} ,\{ 3\} ,\{ 1,2\} ,\{ 1,3\} ,\{ 2,3\} ,\{ 1,2,3\} \} . \]
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Yes
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Proposition 3.3.1. If \( A \) is finite with \( \left| A\right| = n \), then \( \left| {\mathcal{P}\left( A\right) }\right| = {2}^{n} \) .
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Proof. Let\n\n\[ B = \left\{ {\left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) : {b}_{i} = 0\text{ or }{b}_{i} = 1, i = 1,2,\ldots, n}\right\} \]\n\n(3.3.1)\n\nand let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be the elements of \( A \) . Define \( \varphi : B \rightarrow \mathcal{P}\left( A\right) \) by letting\n\n\[ \varphi \left( {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right) = \left\{ {{a}_{i} : {b}_{i} = 1, i = 1,2,\ldots, n}\right\} .\n\n\( \left( {3.3.2}\right) \)\n\nThen \( \varphi \) is a one-to-one correspondence. The result now follows from the next exercise.\n\nQ.E.D.
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No
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Theorem 3.3.2. If \( A \) is a nonempty set, then \( \left| A\right| < \left| {\mathcal{P}\left( A\right) }\right| \) .
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Proof. Define \( \varphi : A \rightarrow \mathcal{P}\left( A\right) \) by \( \varphi \left( a\right) = \{ a\} \) . Then \( \varphi \) is one-to-one. Now suppose \( \psi : A \rightarrow \mathcal{P}\left( A\right) \) is any one-to-one function. Let\n\n\[ C = \{ a : a \in A, a \notin \psi \left( a\right) \} . \]\n\nSuppose there exists \( a \in A \) such that \( \psi \left( a\right) = C \) . Then \( a \in C \) if and only if \( a \notin C \), an obvious contradiction. Thus \( C \) is not in the range of \( \psi \), and so \( \psi \) is not a one-to-one correspondence. Q.E.D.
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Yes
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Lemma 3.3.3. Let \( A \) be the set of all sequences \( {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty } \) with \( {a}_{i} = 0 \) or \( {a}_{i} = 1 \) for each \( i = 1,2,3,\ldots \) Then \( \left| A\right| = \left| {\mathcal{P}\left( {\mathbb{Z}}^{ + }\right) }\right| \) .
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Proof. Define \( \varphi : A \rightarrow \mathcal{P}\left( {\mathbb{Z}}^{ + }\right) \) by\n\n\[ \varphi \left( {\left\{ {a}_{i}\right\} }_{i = 1}^{\infty }\right) = \left\{ {i : i \in {\mathbb{Z}}^{ + },{a}_{i} = 1}\right\} .\n\]\n\nThen \( \varphi \) is a one-to-one correspondence.\n\nQ.E.D.
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Yes
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Proposition 4.1.1. If \( a, b \in \mathbb{R} \) with \( a < b \), then\n\n\[ \left( {a, b}\right) = \{ x : x = {\lambda a} + \left( {1 - \lambda }\right) b,0 < \lambda < 1\} . \]
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Proof. Suppose \( x = {\lambda a} + \left( {1 - \lambda }\right) b \) for some \( 0 < \lambda < 1 \) . Then\n\n\[ b - x = {\lambda b} - {\lambda a} = \lambda \left( {b - a}\right) > 0, \]\n\n(4.1.8)\n\nso \( x < b \) . Similarly,\n\n\[ x - a = \left( {\lambda - 1}\right) a + \left( {1 - \lambda }\right) b = \left( {1 - \lambda }\right) \left( {b - a}\right) > 0, \]\n\n(4.1.9)\n\nso \( a < x \) . Hence \( x \in \left( {a, b}\right) \) .\n\nNow suppose \( x \in \left( {a, b}\right) \) . Then\n\n\[ x = \left( \frac{b - x}{b - a}\right) a + \left( \frac{x - a}{b - a}\right) b = \left( \frac{b - x}{b - a}\right) a + \left( {1 - \frac{b - x}{b - a}}\right) b \]\n\n(4.1.10)\n\nand\n\n\[ 0 < \frac{b - x}{b - a} < 1 \]\n\n(4.1.11)\n\nQ.E.D.
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Yes
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Proposition 4.2.1. Every open interval \( I \) is an open set.
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Proof. Suppose \( I = \left( {a, b}\right) \), where \( a < b \) are extended real numbers. Given \( x \in I \), let \( \epsilon \) be the smaller of \( x - a \) and \( b - x \) . Suppose \( y \in \left( {x - \epsilon, x + \epsilon }\right) \) . If \( b = + \infty \), then \( b > y \) ; otherwise, we have\n\n\[ b - y > b - \left( {x + \epsilon }\right) = \left( {b - x}\right) - \epsilon \geq \left( {b - x}\right) - \left( {b - x}\right) = 0, \]\n\nso \( b > y \) . If \( a = - \infty \), then \( a < y \) ; otherwise,\n\n\[ y - a > \left( {x - \epsilon }\right) - a = \left( {x - a}\right) - \epsilon \geq \left( {x - a}\right) - \left( {x - a}\right) = 0, \]\n\nso \( a < y \) . Thus \( y \in I \) and \( I \) is an open set.\n\nQ.E.D.
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Yes
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Proposition 4.2.2. Suppose \( A \) is a set and, for each \( \alpha \in A,{U}_{\alpha } \) is an open set. Then\n\n\[ \mathop{\bigcup }\limits_{{\alpha \in A}}{U}_{\alpha } \]\n\nis an open set.
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Proof. Let \( x \in \mathop{\bigcup }\limits_{{\alpha \in A}}{U}_{\alpha } \) . Then \( x \in {U}_{\alpha } \) for some \( \alpha \in A \) . Since \( {U}_{\alpha } \) is open, there exists an \( \epsilon > 0 \) such that \( \left( {x - \epsilon, x + \epsilon }\right) \subset {U}_{\alpha } \) . Thus\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \subset {U}_{\alpha } \subset \mathop{\bigcup }\limits_{{\alpha \in A}}{U}_{\alpha } \]\n\nHence \( \mathop{\bigcup }\limits_{{\alpha \in A}}{U}_{\alpha } \) is open.\n\nQ.E.D.
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Yes
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