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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 . | 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 ... | No |
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. | 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... | Yes |
Theorem 5.2.3 There exists a unique hyperbolic line through any two distinct points in the hyperbolic plane. | 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}}_{\i... | Yes |
Theorem 5.2.4 Any two hyperbolic lines are congruent in hyperbolic geometry. | 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 \( {p}^{ * } \)... | Yes |
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 \( L \) . | 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 \... | No |
Theorem 5.2.11 Given any points \( p \) and \( q \) in \( \mathbb{D} \), there exists a hyperbolic circle centered at \( p \) through \( q \) . | 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 \(... | No |
Example 5.3.4 The distance between two points. | 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. | No |
Corollary 5.3.8 All points on a hyperbolic circle centered at \( p \) are equidistant from \( p \) . | 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... | Yes |
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. | 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 \... | Yes |
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}\lef... | 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-negat... | Yes |
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? | 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 describ... | Yes |
The area of a circle in \( \left( {\mathbb{D},\mathcal{H}}\right) \) . | 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\... | Yes |
Theorem 5.4.8 Any ideal triangle has area equal to \( \pi \) . | 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 \( \... | Yes |
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\[ A = \pi - \left( {\alpha + \beta + \gamma }\right) \] | 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... | Yes |
Corollary 5.4.13 Hyperbolic hypotenuse theorem. In a right hyperbolic triangle with hyperbolic side lengths \( a \) and \( b \), and hypotenuse \( c \) , | \[ \cosh \left( c\right) = \cosh \left( a\right) \cosh \left( b\right) . \] | Yes |
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? | 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 tr... | No |
Inscribe a circle in an ideal triangle. | 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... | Yes |
Example 5.5.4 The distance between \( {ri} \) and \( {si} \) . | 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,\in... | Yes |
The area of a \( \frac{2}{3} \) -ideal triangle. | 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 \( \fr... | Yes |
Lemma 6.1.2 Given two diametrically opposed points on the unit sphere, their image points under stereographic projection are antipodal points in \( {\mathbb{C}}^{ + } \) . | 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... | Yes |
Lemma 6.1.4 If a cline in \( {\mathbb{C}}^{ + } \) contains two antipodal points then it is a great circle. | 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 ... | No |
Theorem 6.1.6 Reflection of \( {\mathbb{S}}^{2} \) about a great circle corresponds via stereographic projection to inversion about a great circle in \( {\mathbb{C}}^{ + } \) . | We work through the relationship in one case, and refer the interested reader to \( \left\lbrack {10}\right\rbrack \) for the general proof. | No |
We argue that reflection of \( {\mathbb{S}}^{2} \) about the equator corresponds to inversion about the unit circle. | 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... | Yes |
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}}^{... | 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} +... | Yes |
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}}... | 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 \( \sqrt{3} \). | Yes |
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 lin... | \n\nFigure 6.1.11 Constructing the antipodal point to \( z \) . | No |
Theorem 6.2.4 If a Möbius transformation preserves antipodal points, then it is an elliptic Möbius transformation. | 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 point... | Yes |
Theorem 6.2.11 There is a unique elliptic line connecting two points \( p \) and \( q \) in \( {\mathbb{P}}^{2} \) . | 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 ... | Yes |
Theorem 6.2.12 The set of elliptic lines is a minimally invariant set of elliptic geometry. | 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, ... | Yes |
Theorem 6.2.13 Any two elliptic lines intersect in \( {\mathbb{P}}^{2} \) . | 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 ... | Yes |
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\... | 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 \ | No |
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 \( {2\alpha } \) . | 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![d4f13802-8387-4... | Yes |
Example 6.3.11 Triangle area in \( \left( {{\mathbb{P}}^{2},\mathcal{S}}\right) \) . | 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... | Yes |
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\[ A = \left( {\alpha + \beta + \gamma }\right) - \pi . \] | From this theorem it follows that the angles of any triangle in elliptic geometry sum to more than \( {180}^{ \circ } \) . | No |
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 \( c \) , | \[ \\cos \\left( c\\right) = \\cos \\left( a\\right) \\cos \\left( b\\right) \] | Yes |
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 \( {2\alpha }/k \) . | 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 ... | Yes |
Lemma 7.2.2 In elliptic geometry with curvature \( k \), the area of a triangle with angles \( \alpha ,\beta \), and \( \gamma \) is\n\n\[ A = \frac{1}{k}\left( {\alpha + \beta + \gamma - \pi }\right) \] | 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. | No |
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\[ \tan \left( {\theta /2}\right) = {e}^{-\sqrt{\left| k\right| }d}. \] | 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 th... | Yes |
Theorem 7.4.2 For all real numbers \( k,\left( {{X}_{k},{G}_{k}}\right) \) is homogeneous and isotropic. | 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,... | Yes |
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\[ \n{kA} = \left( {\alpha + \beta + \gamma - \pi }\right) .\n\] | 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... | No |
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 =... | 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 \(... | Yes |
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\[ \frac{{4\pi }{x}^{2}}{1 + k{x}^{2}} \] | 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... | No |
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\le... | 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... | Yes |
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. | Each point in \( S \) has a neighborhood of points that is an open 2-ball. | No |
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 \( {\mathbb{T}}^{2}\# {\mathbb{T}}^{2} \) i... | 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 ho... | Yes |
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. | 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]. | No |
Why is the Klein bottle non-orientable? | 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... | Yes |
Example 7.5.21 Attempted cell divisions of \( {H}_{1} \) . | 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\nBoth of the terms on the right are nonnegative by Exercise 1.3.8. Hence the sum is non... | No |
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 \( {s}^{2} = 2 \) . | 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{{... | Yes |
Proposition 1.3.2. There does not exist a rational number \( s \) with the property that \( {s}^{2} = 2 \) . | 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}^{... | Yes |
We have\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}\frac{1}{i} = 0 \] | 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 \( \frac{1}{\epsilon } \) . | Yes |
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. | 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\... | Yes |
Proposition 1.4.1. For any \( a, b \in \mathbb{R},\left| {a + b}\right| \leq \left| a\right| + \left| b\right| \) . | 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 no... | No |
Proposition 1.4.2. Given \( a \in {\mathbb{R}}^{ + } \), there exist \( r, s \in \mathbb{Q} \) such that \( 0 < r < a < s \) . | 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 \( {u}_{i} - r > \frac{\epsilon }{2} \) f... | Yes |
Proposition 1.4.3. \( \mathbb{R} \) is an archimedean ordered field. | 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... | Yes |
Proposition 1.4.4. Given \( a, b \in \mathbb{R} \) with \( a < b \), there exists \( r \in \mathbb{Q} \) such that \( a < r < b \) . | 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 ... | Yes |
Theorem 1.4.5. Suppose \( A \subset \mathbb{R}, A \neq \varnothing \), has an upper bound. Then \( \sup A \) exists. | 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}... | Yes |
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. | 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 \(... | 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{\li... | 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|... | No |
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... | 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 }}{... | Yes |
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 \( L \in \mathbb{R} \) . | 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 \( i, j > N \) , \( ... | Yes |
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... | 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... | Yes |
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 \( {y}_{i} \neq 0 \) fo... | 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}\ri... | Yes |
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\\[ \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 \... | 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... | 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}} +... | Yes |
Proposition 2.1.10. If \( x \in \mathbb{R} \) and \( \left| x\right| < 1 \), then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n} = 0 \] | 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... | 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 \( \Lambda \neq \varnothing \) . | 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 seq... | Yes |
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\[ \mathop{\limsup }\limits_{{i \rightarrow \infty }}{x}_{i} = \sup \Lambda \] | 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 \i... | Yes |
Proposition 2.2.1. Suppose \( \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i} \) converges. Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . | 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} ... | Yes |
Proposition 2.2.2. For any real number \( x \) with \( \left| x\right| < 1 \) , \n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{x}^{n} = \frac{1}{1 - x} \] | 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 \inft... | Yes |
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 }... | 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 }{... | No |
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 }... | 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... | No |
Consider the infinite series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{3!} + \frac{1}{4!} + \cdots . \] | 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 }\limi... | Yes |
Proposition 2.2.5. \( e \notin \mathbb{Q} \) . | 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 \(... | Yes |
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{\s... | 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. | Yes |
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{\s... | 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. | No |
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 \( n = 1,2,3,\ldots \) | 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 -... | Yes |
Proposition 3.1.2. With the notation as above, \[ x = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{2}^{n}} \] | 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 \ri... | Yes |
Lemma 3.1.3. With the notation as above, given any integer \( N \) there exists an integer \( n > N \) such that \( {a}_{n} = 0 \) . | 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 ... | Yes |
Proposition 3.2.1. Suppose \( A \) and \( B \) are countable sets. Then the set \( C = \) \( A \cup B \) is countable. | 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\{ \beg... | Yes |
Proposition 3.2.2. Suppose \( A \) and \( B \) are countable. Then \( C = A \times B \) is countable. | 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) =... | Yes |
Proposition 3.2.3. \( \mathbb{Q} \) is countable. | 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 \( \l... | Yes |
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. | 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.... | Yes |
If \( A = \{ 1,2,3\} \), then | \[ \mathcal{P}\left( A\right) = \{ \varnothing ,\{ 1\} ,\{ 2\} ,\{ 3\} ,\{ 1,2\} ,\{ 1,3\} ,\{ 2,3\} ,\{ 1,2,3\} \} . \] | Yes |
Proposition 3.3.1. If \( A \) is finite with \( \left| A\right| = n \), then \( \left| {\mathcal{P}\left( A\right) }\right| = {2}^{n} \) . | 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\[ \varp... | No |
Theorem 3.3.2. If \( A \) is a nonempty set, then \( \left| A\right| < \left| {\mathcal{P}\left( A\right) }\right| \) . | 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\nSuppos... | Yes |
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 \( \left| A\right| = \left| {\mathcal{P}\left( {\mathbb{Z}}^{ + }\right) }\right| \) . | 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. | Yes |
Proposition 4.1.1. If \( a, b \in \mathbb{R} \) with \( a < b \), then\n\n\[ \left( {a, b}\right) = \{ x : x = {\lambda a} + \left( {1 - \lambda }\right) b,0 < \lambda < 1\} . \] | 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 -... | Yes |
Proposition 4.2.1. Every open interval \( I \) is an open set. | 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 >... | Yes |
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. | 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 - \e... | Yes |
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