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Theorem 10.5. The degree of an extension \( \mathbb{Q}\left( r\right) \) always matches the degree of the irreducible polynomial to which \( r \) is a root. | For example, \( \mathbb{Q}\left( \sqrt{2}\right) \) is a degree-2 extension and \( \sqrt{2} \) is a root of the degree-2 irreducible polynomial \( {x}^{2} - 2 \) . | Yes |
Theorem 10.7. Successive extensions multiply degrees. | \[ \left\lbrack {\mathbb{Q}\left( {a, b}\right) : \mathbb{Q}}\right\rbrack = \left\lbrack {\mathbb{Q}\left( {a, b}\right) : \mathbb{Q}\left( a\right) }\right\rbrack \left\lbrack {\mathbb{Q}\left( a\right) : \mathbb{Q}}\right\rbrack \] This theorem is best shown in a diagram like Figure 10.10. On the bottom is the field... | Yes |
Theorem 10.8. Take any polynomial irreducible over \( \mathbb{Q} \) and any two of its roots, \( {r}_{1} \) and \( {r}_{2} \). | 1. There is an isomorphism \( \phi : \mathbb{Q}\left( {r}_{1}\right) \rightarrow \mathbb{Q}\left( {r}_{2}\right) \) that replaces every \( {r}_{1} \) with \( {r}_{2} \), but fixes \( \mathbb{Q} \) . For example, in a degree-2 extension, \( \phi \) would look like this.\n\n\[ \phi \left( {a + b{r}_{1}}\right) = a + b{r}... | Yes |
Theorem 10.9. The degree of a normal extension equals the order of its Galois group. | This theorem makes it easier to compute the Galois group of \( \mathbb{Q}\left( {{r}_{1},{r}_{2},{r}_{3}}\right) \) . Because there are three roots, the Galois group (which permutes those roots) must be isomorphic to some subgroup of \( {S}_{3} \) . Theorem 10.9 says that the order of the Galois group must match the de... | Yes |
Lemma 1.5. (Zorn’s Lemma) Suppose a partially ordered set \( \left( {P, \leq }\right) \) has the property that every chain has an upper bound in \( P \) . Then the set \( P \) contains at least one maximal element. | Proof. A nice discussion of the proof and its historical context is given in Sti13, Section 7.2]. Alternatively see also [Cie97, Theorem 4.3.4] and [Je73, p. 9] for a proof. In fact, as is shown in [Sti13, p. 154] or alternatively [Je73, Theorem 2.1], Zorn's Lemma is equivalent to the Axiom of Choice. | No |
Theorem 1.6. Well-ordering Theorem Every set admits a well-order. | Proof. We refer to [Cie97, Theorem 4.3.3] or alternatively to [Je73, p. 9] for a proof. In fact, as for Zorn's Lemma it is shown in [Je73, Theorem 2.1] that the Well-ordering Theorem is equivalent to the Axiom of Choice. | No |
(1) Every subset of a countable set is again countable. | (1) After excluding the case that \( S \) is finite and the case that \( T \) is finite one sees that one has to prove the following: given any infinite subset \( T \) of \( \mathbb{N} \) there exists a bijection \( \varphi : \mathbb{N} \rightarrow T \) . Such a bijection is given by defining \( \varphi \left( i\right)... | No |
Lemma 2.1. Let \( \left( {X, d}\right) \) be a metric space. A subset \( U \) of \( X \) is called open if for every \( x \in U \) there exists an \( \epsilon > 0 \) such that \( {B}_{\epsilon }\left( x\right) \mathrel{\text{:=}} \{ y \in X \mid d\left( {x, y}\right) < \epsilon \} \) is contained in \( U \) . These ope... | Proof.\n\n(1) It is clear that the empty set and \( X \) itself are open.\n\n(2) Suppose \( {U}_{1},\ldots ,{U}_{k} \) are open subsets of \( X \) . We need to show that the intersection \( {U}_{1} \cap {U}_{2} \cap \cdots \cap {U}_{k} \) is also open. Thus let \( x \in {U}_{1} \cap {U}_{2} \cap \cdots \cap {U}_{k} \) ... | Yes |
Proposition 2.3. Let \( \left( {X, d}\right) \) be a metric space, let \( y \in X \) and let \( r \in \mathbb{R} \) .\n\n(1) the set \( {B}_{r}\left( y\right) = \{ x \in X \mid d\left( {x, y}\right) < r\} \) is open in \( X \) , | Proof. Let \( \left( {X, d}\right) \) be a metric space, let \( y \in X \) and let \( r \in \mathbb{R} \) .\n\n(1) Let \( x \in {B}_{r}\left( y\right) \) . We set \( s \mathrel{\text{:=}} d\left( {x, y}\right) \) . By definition of \( {B}_{r}\left( y\right) \) we have \( s < r \) . We set \( t \mathrel{\text{:=}} r - s... | Yes |
Lemma 2.5. Let \( X \) be a topological space and let \( U \subset X \) be a subset. If given any \( x \in U \) there exists a neighborhood \( V \) of \( x \) that is contained in \( U \), then \( U \) is open. | Proof (*). Let \( X \) be a topological space and let \( U \subset X \) be a subset. Suppose that given any \( x \in U \) there exists a neighborhood \( {V}_{x} \) of \( x \) that is contained in \( U \) . By definition of a neighborhood there exists for each \( x \in U \) an open subset \( {W}_{x} \) with \( x \in {W}... | Yes |
Let \( X \) be a topological space and let \( M \) be a subset of \( X \). Let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \). The following two statements hold:\n\n(a) \( M \) is open if and only if each intersection \( M \cap {U}_{i} \) is open in \( {U}_{i} \),\n\n(b) \( M \) is closed if and... | Proof. Let \( X \) be a topological space and let \( M \) be a subset of \( X \).\n\n(1) Let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \). The \ | No |
(1) Every open subset of \( {\mathbb{R}}^{n} \) is the union of open balls of the form \( {B}_{\epsilon }\left( x\right) \) where \( \epsilon \) is rational and \( x \in {\mathbb{Q}}^{n} \) . | Proof \( \left( *\right) \) .\n\n(1) Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \) . For \( x \in U \) we set \( {\mu }_{x} \mathrel{\text{:=}} \sup \left\{ {\mu \in \left\lbrack {0,1}\right\rbrack \mid {B}_{\mu }\left( x\right) \subset U}\right\} \) . The hypothesis that \( U \) is open implies that \( {\mu ... | Yes |
Lemma 2.8. Let \( X \) be a topological space and let \( A \) be a subset of \( X \). (1) The interior \( \overset{ \circ }{A} \) is an open subset of \( X \). | Proof. (1) The union of arbitrarily many open sets is, by definition of a topology, again an open set. Thus the interior \( \overset{ \circ }{A} \) is an open subset of \( X \). | Yes |
Lemma 2.9. Let \( X \) be a topological space and let \( A \) be a subset of \( X \) . We have the following three equalities:\n\n(1) \( \;\mathring{A} = \{ x \in A \mid \) there exists a neighborhood \( U \) of \( x \) that is contained in \( A\} \)\n\n(2) \( \;\bar{A} = \{ x \in X \mid \) every neighborhood of \( x \... | Proof. The proof of the lemma is Exercise 2.3. | No |
Lemma 2.10. (*) Let \( X \) be a topological space and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be a family of subsets. The following inclusions hold:\n\n1) interior of \( \mathop{\bigcup }\limits_{{i \in I}}{U}_{i} \supset \mathop{\bigcup }\limits_{{i \in I}}{\overset{ \circ }{U}}_{i}\; \) and\n\n(2)\n\nFurtherm... | Proof. The statements follow fairly easily from the definitions. We leave it to the reader to verify the statements. | No |
Proposition 2.11. Let \( \left( {X, d}\right) \) be a metric space. The corresponding topological space is Hausdorff. | Proof. Let \( x, y \in X \) be two different points. We set \( \epsilon \mathrel{\text{:=}} \frac{1}{2}d\left( {x, y}\right) \) . It follows from Proposition 2.3 that \( U \mathrel{\text{:=}} {B}_{\epsilon }\left( x\right) \) and \( V \mathrel{\text{:=}} {B}_{\epsilon }\left( y\right) \) are neighborhoods of \( x \) re... | Yes |
Lemma 2.13. Let \( X \) be a topological space. If \( X \) is Hausdorff, then for every \( x \in X \) the corresponding subset \( \{ x\} \) is closed. | Proof. Let \( x \in X \) . Since \( X \) is Hausdorff we know that for every \( y \neq x \) there exist open disjoint neighborhoods \( {U}_{y} \) of \( x \) and \( {V}_{y} \) of \( y \) . We have\n\n\[ \nX \smallsetminus \{ x\} = \mathop{\bigcup }\limits_{{y \in X\smallsetminus \{ x\} }}\{ y\} \underset{ \uparrow }{ \s... | Yes |
Lemma 2.14. Let \( X \) be a topological space. If \( X \) is Hausdorff, then every convergent sequence in \( X \) has a unique limit. | Proof. Assume there exists a sequence \( {\left\{ {a}_{n}\right\} }_{n \in \mathbb{N}} \) in \( X \) that converges to two different points \( x \) and \( y \) . Since \( X \) is Hausdorff there exist two disjoint neighborhoods \( U \) and \( V \) of \( x \) and \( y \) . But by our hypothesis, for large enough \( n \)... | Yes |
Lemma 2.15. Let \( X \) be a topological space, let \( A \subset X \) be a subset and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( A \) . If \( A \) is compact, then there exist finitely many indices \( {i}_{1},\ldots ,{i}_{k} \in I \) such that\n\n\[ A \subset {U}_{{i}_{1}} \cup \cdots \cup {U... | Proof \( \left( *\right) \) . Given \( i \in I \) we set \( {V}_{i} \mathrel{\text{:=}} {U}_{i} \cap A \) . By definition of the subspace topology each \( {V}_{i} \) is an open subset of \( A \) . Thus \( {\left\{ {V}_{i}\right\} }_{i \in I} \) is an open cover of \( A \) . Since \( A \) is compact there exist \( {i}_{... | Yes |
Lemma 2.16. Let \( X \) be a topological space.\n\n(1) The union of finitely many compact subsets of \( X \) is again a compact subset.\n\n(2) The intersection of arbitrarily many compact subsets of \( X \) is again a compact subset. | Proof. The statement follows easily from the definitions, we leave it to the reader to fill in the details. | No |
(1) Let \( X \) be a topological space. If \( X \) is compact, then any closed subset of \( X \) is also compact. | (1) Let \( X \) be a compact topological space and let \( A \) be a closed subset. Let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( A \) . By definition of the subset topology there exists for every \( i \in I \) an open subset \( {V}_{i} \) of \( X \) such that \( {U}_{i} = {V}_{i} \cap A \) . Sin... | Yes |
(1) Every discrete topological space that is compact is finite. | (1) Let \( X \) be a discrete topological space. Each point thus defines an open set. Evidently \( X \) is covered by all the open sets given by the points in \( X \) . Since \( X \) is compact we can cover \( X \) by finitely many of those open sets. But this means that \( X \) contains only finitely many points. | Yes |
Lemma 2.19. Let \( A \subset {\mathbb{R}}^{n} \) be a subset. If \( A \) is compact, then \( A \) is closed and it is bounded. | Proof. Let \( A \subset {\mathbb{R}}^{n} \) be a compact subset.\n\n(1) We want to show that \( {\mathbb{R}}^{n} \smallsetminus A \) is open. Let \( x \in {\mathbb{R}}^{n} \smallsetminus A \) . We need to show that there exists an \( \epsilon > 0 \) with \( {B}_{\epsilon }^{n}\left( x\right) \subset {\mathbb{R}}^{n} \s... | Yes |
Lemma 2.22. Let \( X \) be a set and let \( \mathcal{B} \subset \mathcal{P}\left( X\right) \). We define\n\n\( \mathcal{T}\left( \mathcal{B}\right) \mathrel{\text{:=}} \{ V \subset X \mid \) for every \( x \in V \) there exists a \( B \in \mathcal{B} \) with \( x \in B \subset V\} .\n\nIf \( \mathcal{B} \) has the basi... | Proof (*). Let \( X \) be a set and let \( \mathcal{B} \subset \mathcal{P}\left( X\right) \). We write \( \mathcal{T} \mathrel{\text{:=}} \mathcal{T}\left( \mathcal{B}\right) \). We need to show that \( \mathcal{T} \) satisfies the three properties (1),(2) and (3) in the definition of a topological space, see page 83. ... | Yes |
Lemma 2.23. Let \( X \) be a set and let \( \mathcal{S} \) and \( \mathcal{T} \) be topologies for \( X \) . Suppose that \( \mathcal{C} \) is a basis for \( \mathcal{T} \) . (As discussed above we could take \( \mathcal{C} = \mathcal{T} \) .) If given any \( x \in X \) and given any \( V \in \mathcal{C} \) there exist... | Proof. It follows immediately from Lemma 2.5 that any set in \( \mathcal{C} \) is open in \( \mathcal{S} \) . But then, once again using Lemma 2.5 we see that any set in \( \mathcal{T} \) is open in \( \mathcal{S} \), i.e. \( \mathcal{T} \subset \mathcal{S} \) . | No |
Lemma 2.24. Let \( X \) be a set and let \( \mathcal{B} \subset \mathcal{P}\left( X\right) \). If \( \mathcal{B} \) has the basis property, then\n\n\( V \subset X \) is open with respect to \( \mathcal{T}\left( \mathcal{B}\right) \Leftrightarrow V \) is the union of sets in \( \mathcal{B} \). | Proof of Lemma 2.24 (*) . We first prove the \ | No |
Lemma 2.25. Let \( \mathcal{B} \) be a basis for the topology of a topological space \( X \). Furthermore let \( {\left\{ {U}_{j}\right\} }_{j \in J} \) be an open cover of \( X \), i.e. \( {\left\{ {U}_{j}\right\} }_{j \in J} \) is a family of open sets with \( \mathop{\bigcup }\limits_{{j \in J}}{U}_{j} = X \). Then\... | Proof \( \left( *\right) \). We write\n\n\[ \mathcal{C} \mathrel{\text{:=}} \left\{ {B \in \mathcal{B} \mid \text{ there exists some }j \in J\text{ with }B \subset {U}_{j}}\right\} .\n\nSince \( \mathcal{C} \subset \mathcal{B} \) we have \( \mathcal{T}\left( \mathcal{C}\right) \subset \mathcal{T}\left( \mathcal{B}\righ... | Yes |
Lemma 2.26. Let \( X \) be a set. For any \( \mathcal{C} \subset \mathcal{P}\left( X\right) \) the set\n\n\[ \mathcal{B}\left( \mathcal{C}\right) \mathrel{\text{:=}} \overset{\text{ all subsets of }X\text{ that can be written as }}{\text{ }\text{ all subsets of }\text{ }X\text{ that can be written as }\text{ }}\]\n\nha... | Proof \( \left( *\right) \) .\n\n(1) On page 72 we pointed out that if we take the empty family of subsets of \( X \), then the intersection is all of \( X \) . In other words, we see that \( X \in \mathcal{B}\left( \mathcal{C}\right) \) . It follows that \( \mathcal{B}\left( \mathcal{C}\right) \) satisfies (B1).\n\n(2... | No |
Lemma 2.27. Let \( X \) be a topological space and let \( \mathcal{C} \) be a family of open sets.\n\n(1) Suppose \( \mathcal{C} \) has the following property:\n\n\( \left( *\right) \) Given any open set \( U \) and given any \( x \in U \) there exists a \( C \in \mathcal{C} \) with \( x \in C \subset U \) . Then \( \m... | Proof \( \left( *\right) \) .\n\n(1) We denote by \( \mathcal{T} \) the topology of \( X \) . First we show that \( \mathcal{C} \) has the basis property:\n\n(B1) If we apply the property of \( \mathcal{C} \) to \( U = X \) then we obtain immediately that given any \( x \in X \) there exists a \( C \in \mathcal{C} \) w... | Yes |
Lemma 2.28. (*) Let \( X \) be a topological space and let \( A \subset X \) be a subset. If \( \mathcal{B} \) is a basis for the topology of \( X \), then \( \{ B \cap A \mid B \in \mathcal{B}\} \) is a basis for the topology of \( A \), equipped with the subspace topology. | Proof (*). We prove the lemma using Lemma 2.27 (1). Thus let \( U \subset A \) be an open subset in \( A \) and let \( x \in U \) . By definition of the subspace topology there exists an open subset \( V \subset X \) of \( X \) with \( U = V \cap A \) . Since \( \mathcal{B} \) is a basis for the topology of \( X \) the... | Yes |
(1) If \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) are two continuous maps between topological spaces, then the composition \( g \circ f : X \rightarrow Z \) is also continuous. | (1) Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) be two continuous maps between topological spaces. We need to show that the composition \( g \circ f : X \rightarrow Z \) is also continuous. Let \( U \subset Z \) be an open subset. Then\n\n\[
{\left( g \circ f\right) }^{-1}\left( U\right) = {f}^{-1}\left... | Yes |
Lemma 2.31. Let \( X \) be a topological space. If \( f : X \rightarrow \mathbb{R} \) and \( g : X \rightarrow \mathbb{R} \) are continuous maps, then the maps\n\n\( X \rightarrow \mathbb{R}\;X \rightarrow \mathbb{R}\;X \rightarrow \mathbb{R}\;X \rightarrow \mathbb{R} \)\n\n\( x \mapsto f\left( x\right) + g\left( x\rig... | Proof. In Exercise 2.27 we will show that the maps\n\n\[ \begin{aligned} \alpha : {\mathbb{R}}^{2} & \rightarrow \mathbb{R} \\ \left( {x, y}\right) & \mapsto x + y \end{aligned}\;\text{ and }\;\begin{aligned} \beta : {\mathbb{R}}^{2} & \rightarrow \mathbb{R} \\ \left( {x, y}\right) & \mapsto x \cdot y \end{aligned} \]\... | No |
Lemma 2.32. A map \( f : X \rightarrow Y \) between two topological spaces \( X \) and \( Y \) is continuous if and only if for each closed set \( A \) in \( Y \) the preimage \( {f}^{-1}\left( A\right) \) is closed in \( X \) . | Proof. This lemma is a straightforward consequence of Lemma 1.3 (7) and the definitions. We leave it to the reader to fill in the details. | No |
Lemma 2.34. Let \( X \) be a topological space and let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a family of subsets. The following two statements are equivalent:\n\n(1) A subset \( U \subset X \) is open if each \( U \cap {X}_{i} \) is open in \( {X}_{i} \) .\n\n(2) A subset \( U \subset X \) is closed if each \( ... | Proof.\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) We assume that (1) holds. Now let \( U \subset X \) be a subset. We have\n\n\[ \text{since}\left( {X \smallsetminus U}\right) \cap {X}_{i} = {X}_{i} \smallsetminus \left( {{X}_{i} \cap U}\right) \]\n\n\( \Rightarrow X \smallsetminus U \) is open in \( X \Right... | No |
Lemma 2.35. (Pasting Lemma) Let \( f : X \rightarrow Y \) be a map between topological spaces.\n\n(1) If there exists an open subsets \( {\left\{ {W}_{i}\right\} }_{i \in I} \) of \( X \) such that each restriction \( {\left. f\right| }_{{W}_{i}} \) is continuous, then \( f \) itself is continuous.\n\n(2) If there exis... | Proof. Let \( f : X \rightarrow Y \) be a map between topological spaces.\n\n(1) This statement follows immediately from the combination of Lemma 2.6 (1a) and Lemma 2.34\n\n(2) Similar to (1) this statement follows immediately from the combination of Lemma 2.6 (2a) and Lemma 2.34. | No |
Proposition 2.37. Let \( f : X \rightarrow Y \) be a map between topological spaces. Let \( \mathcal{C} \) be a subbasis for the topology of \( Y \) (e.g. \( \mathcal{C} \) could be a basis for the topology). Then the following holds\n\n\( f \) is continuous \( \Leftrightarrow \) for each \( C \in \mathcal{C} \) the pr... | Proof (*). Let \( f : X \rightarrow Y \) be a map between topological spaces and let \( \mathcal{C} \) be a subbasis for the topology of \( Y \) . As in Lemma 2.26 we denote by \( \mathcal{B} \mathrel{\text{:=}} \mathcal{B}\left( \mathcal{C}\right) \) all subsets of \( Y \) that can be obtained by intersecting finitely... | No |
Lemma 2.38. Let \( X \) be a topological space.\n\n(1) If \( U \subset X \) is an open subset, then the inclusion map \( i : U \rightarrow X \) is open.\n\n(2) If \( A \subset X \) is a closed subset, then the inclusion map \( i : A \rightarrow X \) is closed. | Proof.\n\n(1) Suppose that \( V \) is an open subset of \( U \) . By definition of the subspace topology there exists an open subset \( W \subset X \) with \( V = U \cap W \) . But then \( i\left( V\right) = V = U \cap W \) is the intersection of two open subsets of \( X \), thus it is an open subset of \( X \) .\n\n(2... | No |
Lemma 2.39. Let \( f : X \rightarrow Y \) be a map between topological spaces. Let \( \mathcal{C} \) be a subbasis for the topology of \( X \) (e.g. \( \mathcal{C} \) could be a basis for the topology). Then the following holds\n\n\( f \) is open \( \Leftrightarrow \) for each \( C \in \mathcal{C} \) the image \( f\lef... | Proof. The logic of the proof is very similar to the proof of Proposition 2.37, Eventually the proof boils down to Lemmas 2.24 and 1.3 . | No |
(1) Let \( f : X \rightarrow Y \) be a continuous map. If \( X \) is compact, then \( f\left( X\right) \) is also compact. | (1) Let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( f\left( X\right) \) . We need to show that \( f\left( X\right) \) is contained in the union of finitely many \( {U}_{i} \) ’s. We have\n\n\[ X\underset{ \uparrow }{ = }{f}^{-1}\left( {f\left( X\right) }\right) \subset {f}^{-1}\left( {\mathop{\big... | Yes |
Lemma 2.41. (*) Let \( f : K \rightarrow X \) be a continuous map from a compact topological space \( K \) to a topological space \( X \) 22 Suppose there exists a sequence \( {U}_{1},{U}_{2},\ldots \) of subsets of \( X \) with the following properties:\n\n(1) each \( {U}_{i} \) is open,\n\n(2) the sequence is nested,... | Proof (*). For \( i \in \mathbb{N} \) we write \( {V}_{i} = {f}^{-1}\left( {U}_{i}\right) \) . Since \( f \) is continuous it follows from (1) that each \( {V}_{i} \) is an open subset of \( K \) . By (3) the \( {V}_{i} \) cover all of \( K \) . Since \( K \) is compact there exist \( {i}_{1},\ldots ,{i}_{n} \) such th... | Yes |
Lemma 2.42. (*) Let \( f : X \rightarrow Y \) be a map between topological spaces that is continuous. We assume that \( f \) is open or that \( f \) is closed.\n\n(1) If \( f \) is an injection, then \( f \) is an open (respectively closed) embedding.\n\n(2) If \( f \) is a bijection, then \( f \) is a homeomorphism. | Proof (*). Let \( f : X \rightarrow Y \) be a map between topological spaces that is continuous\n\n(2) We assume that \( f \) is a bijection. Furthermore we now assume that \( f \) is open. We want to show that \( f \) is a homeomorphism. It remains to show that \( g \mathrel{\text{:=}} {f}^{-1} : Y \rightarrow X \) is... | Yes |
Proposition 2.43. Let \( f : X \rightarrow Y \) be a continuous map between topological spaces. If \( X \) is compact and if \( Y \) is Hausdorff, then the following statements hold:\n\n(1) The map \( f : X \rightarrow Y \) is closed.\n\n(2) If \( f \) is an injection, then \( f \) is a closed embedding.\n\n(3) If \( f... | Proof. Let \( f : X \rightarrow Y \) be a continuous map between topological spaces. We assume that \( X \) is compact and that \( Y \) is Hausdorff.\n\n(1) Let \( A \) be a closed subset of \( X \) . We need to show that \( f\left( A\right) \) is a closed subset of \( Y \) . In fact we see that\n\n\( A \subset X \) cl... | No |
Lemma 2.44. Let \( n \in \mathbb{N} \) . We consider the map\n\n\[ \Phi : {S}^{n} \rightarrow {\mathbb{R}}^{n} \cup \{ \infty \} \]\n\n\[ \left( {{x}_{1},\ldots ,{x}_{n + 1}}\right) \mapsto \left\{ \begin{array}{ll} \left( {\frac{{x}_{1}}{1 - {x}_{n + 1}},\ldots ,\frac{{x}_{n}}{1 - {x}_{n + 1}}}\right) , & \text{ if }{... | Proof.\n\n(1) This statement follows immediately from the definition of \( \Phi \) .\n\n(2) This statement follows from an elementary calculation which we leave to the reader.\n\n(3) This statement follows immediately from the definition of \( \Phi \) .\n\n(4) We will prove this statement in Exercise 2.51.\n\n(5) We le... | No |
Proposition 2.45. Let \( X \) be a topological space, let \( n \in \mathbb{N} \) and let \( f : X \rightarrow {\mathbb{R}}^{n} \) be an injective continuous map. Furthermore suppose that \( X \) is the union of compact subsets \( {\left\{ {K}_{i}\right\} }_{i \in I} \) such that the following condition is satisfied:\n\... | Proof. By Lemma 2.42 (1) it remains to show that the map \( f : X \rightarrow f\left( X\right) \) is closed. In fact, by definition of the subspace topology we only need to show that the map \( f : X \rightarrow {\mathbb{R}}^{n} \) is closed.\n\nThus let \( A \) be a closed subset of \( X \) . We need to show that \( f... | Yes |
Lemma 2.46. (*) Let \( X \) be a Hausdorff space.\n\n(1) Let \( A \) be a subset of \( X \) and let \( B \) be a compact subset of \( X \) . There exists an open subset \( U \) of \( X \) with \( A \subset U \) and \( U \cap B = \varnothing \) .\n\n(2) If \( X \) is compact, then it is also normal. | Proof \( \\left( *\\right) \) . Let \( X \) be a Hausdorff space. We start out with the following claim.\n\nClaim. Let \( A \) be a subset of \( X \) and let \( B \) be a compact subset of \( X \) . Given any \( a \in A \) there exists an open neighborhood \( {U}_{a} \) of \( a \) and an open neighborhood \( {V}_{a} \)... | Yes |
Proposition 2.47. Let \( \left( {X, d}\right) \) be a metric space. If we equip \( X \) with the corresponding topology, then \( X \) is normal. | Proof. We leave the proof of this proposition as a challenging exercise to the reader. Alternatively we refer to [Wil70, Theorems 20.9 and 20.10] for a proof. | No |
Lemma 2.48. Let \( X \) be a topological space and let \( A \subset X \) be a closed subset and \( U \subset X \) be an open subset with \( A \subset U \) . If \( X \) is normal, then there exists an open subset \( V \) with \( A \subset V \subset \bar{V} \subset U \) . | Proof. Let \( X \) be a topological space and let \( A \subset X \) be a closed subset and \( U \subset X \) be an open subset with \( A \subset U \) . We set \( B \mathrel{\text{:=}} X \smallsetminus U \) . This is a closed subset with \( A \cap B = \varnothing \) . Since \( X \) is normal there exist disjoint neighbo... | Yes |
Lemma 2.49. Let \( X \) be a topological space and let \( {U}_{1},\ldots ,{U}_{m} \) be open subsets such that \( X = \mathop{\bigcup }\limits_{{i = 1}}^{m}{U}_{i} \) . If \( X \) is normal, then there exist open subsets \( {V}_{i} \subset {U}_{i} \) such that \( {\bar{V}}_{i} \subset {U}_{i} \) and such that \( X = \m... | Proof \( \left( *\right) \) . To simplify the notation we only discuss the case \( m = 2 \) . We leave it to the reader to prove the general case.\n\nThus let \( X \) be a topological space that is normal. Furthermore let \( {U}_{1} \) and \( {U}_{2} \) be two open subsets with \( {U}_{1} \cup {U}_{2} = X \) . This dat... | No |
Lemma 2.50. (Urysohn’s Lemma) Let \( X \) be a topological space. If \( X \) is normal, then for any two disjoint closed subsets \( A \) and \( B \) there exists a continuous function \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) with \( {\left. f\right| }_{A} \equiv 0 \) and \( {\left. f\right| }_{B} \equiv ... | Proof. Proofs of Urysohn's Lemma are provided in many textbooks on general topology, see e.g. [Kel75, Lemma 4.4], [Mun75, Theorem 4.3.1] or [Jä05, p. 109]. | No |
Theorem 2.51. (Tietze Extension Theorem) Let \( X \) be a topological space, let \( A \subset X \) be a closed subset and let \( f : A \rightarrow \mathbb{R} \) be a continuous map. If \( X \) is normal, then there exists a continuous map \( g : X \rightarrow \mathbb{R} \) with \( {\left. g\right| }_{A} = f \) . | Proof. We refer to [Kel75, p. 242], [Mun75, Theorem 4.3.2] or [Jä05, p. 114] for proofs. | No |
Proposition 2.55. The interval \( \left\lbrack {0,1}\right\rbrack \) is connected. | Proof. Let \( U \) and \( V \) be two disjoint open subsets of \( \left\lbrack {0,1}\right\rbrack \) with \( U \cup V = \left\lbrack {0,1}\right\rbrack \) . Without loss of generality we can assume that \( 0 \in U \) . We want to show that \( U = \left\lbrack {0,1}\right\rbrack \) . We consider\n\n\[ A \mathrel{\text{:... | Yes |
Corollary 2.56. Every path-connected topological space is also connected. | Proof. Let \( X \) be a topological space that is path-connected. Suppose that \( X \) is not connected. This means that there exist disjoint open non-empty subsets \( U \) and \( V \) with \( X = U \cup V \) . Since \( U \) and \( V \) are non-empty we can find points \( x \in U \) and \( y \in V \) . Since \( X \) is... | Yes |
Lemma 2.57. Let \( f : X \rightarrow Y \) be a map between topological spaces.\n\n(1) If \( X \) is path-connected, then \( f\left( X\right) \) is also path-connected.\n\n(2) If \( X \) is connected, then \( f\left( X\right) \) is also connected. | Proof. We leave the elementary proof to the reader. | No |
(1) Let \( n \geq 1 \) . Every open ball \( {B}_{r}^{n}\left( y\right) \), every closed ball \( {\bar{B}}_{r}^{n}\left( y\right) \) and all of \( {\mathbb{R}}^{n} \) are path-connected and connected. | Proof. By Corollary 2.56 it suffices to show that all these topological spaces are path-connected. It is clear that convex subsets of \( {\mathbb{R}}^{n} \) are path-connected.\n\n(1) It follows from the example on page 124 that all balls (open or closed) are path-connected. | Yes |
(1) The topological spaces \( \mathbb{R} \) and \( {\mathbb{R}}^{n} \) are not homeomorphic. | Let us suppose that there exists a homeomorphism \( f : \mathbb{R} \rightarrow {\mathbb{R}}^{n} \) . Let \( P \in \mathbb{R} \) be a point. Then \( f \) restricts to a homeomorphism from \( \mathbb{R} \smallsetminus \{ P\} \) to \( {\mathbb{R}}^{2} \smallsetminus \{ f\left( P\right) \} \) . The topological space \( \ma... | Yes |
Proposition 2.60. Let \( n \in \mathbb{N} \) .\n\n(1) There exists a \( {ma}{p}^{37}f : \left\lbrack {0,1}\right\rbrack \rightarrow {\left\lbrack 0,1\right\rbrack }^{n} \) that is surjective.\n\n(2) There exists an injective map \( f : {S}^{1} \rightarrow {\left\lbrack 0,1\right\rbrack }^{2} \) such that \( f\left( {S}... | Proof.\n\n(1) Many examples of such maps are given in [Sag94], see e.g. [Sag94, Theorem 2.1].\n\n(2) In 1903 William Osgood [Osg03] and Henri Lebesgue [Leb1903] independently showed that there exist injective maps \( f : \left\lbrack {0,1}\right\rbrack \rightarrow {\left\lbrack 0,1\right\rbrack }^{2} \) such that \( f\... | No |
Lemma 2.61. Every map \( f : X \rightarrow Y \) from a connected topological space \( X \) to a discrete topological space is constant. | Proof. We need to show that for any \( x \in X \) we have \( f\left( x\right) = f\left( X\right) \) . Since \( Y \) is a discrete topological space we know that \( \{ f\left( x\right) \} \) and \( Y \smallsetminus \{ f\left( x\right) \} \) are open subsets of \( Y \) . The preimages \( U \mathrel{\text{:=}} {f}^{-1}\le... | Yes |
Lemma 2.62. (*) Let \( X \) be a topological space and let \( {\left\{ {B}_{i}\right\} }_{i \in I} \) be a family of subsets such that the intersection \( \mathop{\bigcap }\limits_{{i \in I}}{B}_{i} \) is non-empty.\n\n(1) If each \( {B}_{i}, i \in I \) is path-connected, then \( \mathop{\bigcup }\limits_{{i \in I}}{B}... | Proof (*). Let \( X \) be a topological space and let \( {\left\{ {B}_{i}\right\} }_{i \in I} \) be a family of subsets such that the intersection \( \mathop{\bigcap }\limits_{{i \in I}}{B}_{i} \) contains a point \( z \) . We write \( Z \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{i \in I}}{B}_{i} \) .\n\n(1) We sup... | Yes |
Lemma 2.63. (*) Let \( X \) be a topological space and let \( A \subset X \) be a subset.\n\n(1) If \( X \) is path-connected and if \( \partial A \) is path-connected, then \( X \smallsetminus \mathring{A} \) is also path-connected.\n\n(2) If \( X \) is connected and if \( \partial A \) is connected, then \( X \smalls... | We will prove Lemma 2.63 in Exercise 2.54. | No |
For every topological space \( X \) the notion of path-equivalence is indeed an equivalence relation. | All properties of an equivalence relation are trivial except possibly for transitivity. The fact that path-equivalence is transitive follows immediately from the argument in Lemma 2.62 (1). | No |
Lemma 2.65. Let \( n \in \mathbb{N} \). (1) The real general linear group \[ \mathrm{GL}\left( {n,\mathbb{R}}\right) = \{ A \in \mathrm{M}\left( {n \times n,\mathbb{R}}\right) \mid \det \left( A\right) \neq 0\} \subset \mathrm{M}\left( {n \times n,\mathbb{R}}\right) = {\mathbb{R}}^{{n}^{2}} \] has precisely two path-co... | Proof. It is an instructive exercise in linear algebra to prove this lemma, see Exercise 2.60, Alternatively, the proof for these statements is given in Bak02, Chapter 9.2]. | No |
For every topological space \( X \) the above notion of equivalence is indeed an equivalence relation. | Proof. All properties of an equivalence relation are trivial except for transitivity. So suppose that \( x \) and \( y \) are equivalent and that \( y \) and \( z \) are equivalent. By definition there exists a connected subset \( A \) that contains \( x \) and \( y \) and there exists a connected subset \( B \) that c... | Yes |
Let \( X \) be a topological space.\n\n(1) As a set \( X \) is the disjoint union of its components.\n\n(2) Each component of \( X \) is connected.\n\n(3) Each component \( C \) of \( X \) is a maximal connected subset of \( X \), i.e. if \( C \subsetneq D \), then \( D \) is not connected.\n\n(4) If \( A \) is a subse... | Proof (*). Let \( X \) be a topological space.\n\n(1) This statement is clear.\n\n(2) Now let \( C \) be a component of \( X \) . We pick \( c \in C \) . For each \( d \in C \) there exists by definition a connected subset \( {Y}_{d} \) of \( X \) that contains \( c \) and \( d \) . We have\n\n\[ C = \mathop{\bigcup }\... | No |
Every component of \( X \) is a closed subset of \( X \) . | Let \( C \) be a component of \( X \) . We need to show that \( X \smallsetminus C \) is open. By Lemma 2.5 it suffices to show that any \( P \in X \smallsetminus C \) admits a neighborhood that does not intersect \( C \) . Thus let \( P \in X \smallsetminus C \) . By Lemma 2.68 we know that \( C \cup \{ P\} \) is not ... | Yes |
Lemma 2.70. (*) Let \( X \) and \( Y \) be topological space. If \( X \) is (path-) connected and if \( Y \) consists of (path-) components \( {Y}_{i}, i \in I \), then given any map \( f : X \rightarrow Y \) there exists an \( i \in I \) with \( f\left( X\right) \subset {Y}_{i} \) . | Proof \( \left( *\right) \) . Again we leave the proof as a straightforward exercise to the reader. | No |
Lemma 2.71. Let \( P \) be a property of topological spaces and let \( X \) be a topological space. If \( X \) is locally \( P \), then the open subsets of \( X \) that have \( P \) form a basis for the topology of \( X \). | Proof. This lemma follows immediately from Lemma 2.27 (1). | No |
Lemma 2.72. Let \( X \) be a topological space that is locally path-connected. Then \( X \) is connected \( \Leftrightarrow X \) is path-connected. | Proof. The \ | No |
Lemma 2.73. (*) Let \( X \) be a Hausdorff space.\n\n(1) If \( X \) is compact, then \( X \) is also regionally compact.\n\n(2) If given any point \( P \in X \) there exists a compact neighborhood of \( P \), then \( X \) is regionally compact. | Proof.\n\n(1) We will use Lemma 2.73 only once, in a rather inessential way. Therefore we leave the proof of this lemma as a not entirely trivial exercise to the reader. Alternatively we refer to [Wil70, Theorem 18.2] for a proof.\n\n(2) This statement follows almost immediately from (1). | No |
Lemma 2.74. Let \( f : X \rightarrow Y \) be a map between two topological spaces (for once we do not assume that \( f \) is continuous). If \( f \) is locally continuous, then \( f \) itself is continuous. | Proof. Let \( f : X \rightarrow Y \) be a locally continuous map between two topological spaces. It follows from Lemma 2.30 (2) that given any \( x \in X \) that there exists an open neighborhood \( {U}_{x} \) of \( x \) such that the restriction of \( f \) to \( {U}_{x} \) is continuous. It is an immediate consequence... | No |
Lemma 2.75. (Lebesgue Lemma) Let \( K \) be a compact metric space and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( K \) . Then there exists a \( \delta > 0 \) such that for every subset \( A \) with \( \operatorname{diam}\left( A\right) < \delta \) there exists an \( i \in I \) with \( A \subs... | Proof. Since \( K \) is compact we can cover \( K \) with finitely many of the \( {U}_{i} \) ’s. Put differently, without loss of generality we can assume that \( I = \{ 1,\ldots, n\} \) is a finite set.\n\nIf there exists an \( i \in \{ 1,\ldots, n\} \) with \( K = {U}_{i} \), then any \( \delta > 0 \) has the desired... | Yes |
Corollary 2.76. Let \( f : {\left\lbrack 0,1\right\rbrack }^{n} \rightarrow X \) be a map from the cube \( {\left\lbrack 0,1\right\rbrack }^{n} \) to a topological space \( X \) and let \( {\left\{ {V}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . Then there exists an \( N > 0 \) such that for any \( {a}_{1}... | Proof. Let \( f : {\left\lbrack 0,1\right\rbrack }^{n} \rightarrow X \) be a map from the cube \( {\left\lbrack 0,1\right\rbrack }^{n} \) to a topological space and let \( {\left\{ {V}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . By Proposition 2.21 we know that \( {\left\lbrack 0,1\right\rbrack }^{n} \) is... | Yes |
Proposition 2.77. Let \( f : X \rightarrow Y \) be a continuous map between two metric spaces. If \( X \) is compact, then \( f \) is uniformly continuous. | Proof (*). Let \( f : X \rightarrow Y \) be a continuous map between two metric spaces and let \( \epsilon > 0 \) . The \ | No |
Lemma 2.78. Let \( X \) be a paracompact topological space and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . There exists a partition of unity \( {\left\{ {f}_{i} : X \rightarrow \left\lbrack 0,1\right\rbrack \right\} }_{i \in I} \) such that for every \( i \in I \) we have \( \operatorna... | Proof. Since \( X \) is paracompact we know, by definition, that there exists a partition of unity \( {\left\{ {g}_{j} : X \rightarrow \left\lbrack 0,1\right\rbrack \right\} }_{j \in J} \) such that for each \( j \in J \) there exists an \( i\left( j\right) \in I \) with \( \operatorname{supp}\left( {g}_{j}\right) \sub... | Yes |
Theorem 2.79. Let \( X \) be a topological space that is Hausdorff. The following two statements are equivalent:\n\n(1) The topological space \( X \) is paracompact.\n\n(2) Every open cover \( {\left\{ {U}_{i}\right\} }_{i \in I} \) admits a locally finite open cover \( {\left\{ {V}_{j}\right\} }_{j \in J} \) that is a... | Proof. The \ | No |
Proposition 2.80. (*) Let \( X \) be a topological space and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . If \( X \) is paracompact, then there exists a countable locally finite open cover \( {\left\{ {V}_{j}\right\} }_{j \in J} \) such that each component of any \( {V}_{j} \) is contain... | Proof (*). Since \( X \) is paracompact there exists a partition of unity \( {\left\{ {f}_{j} : X \rightarrow \left\lbrack 0,1\right\rbrack \right\} }_{j \in J} \) subordinate to the given open cover \( {\left\{ {U}_{i}\right\} }_{i \in I} \) . Given a finite subset \( S \subset J \) we define\n\n\[ \n{V}_{S} \mathrel{... | Yes |
Proposition 2.81. Let \( X \) be a topological space. If \( X \) is compact and Hausdorff, then it is also paracompact. | Proof. Let \( X \) be a topological space that is compact and Hausdorff. Furthermore let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . We perform the following steps:\n\n(1) Since \( X \) is Hausdorff we know by Lemma 2.13 that any subset of \( X \) consisting of a single point is closed.\n\n... | Yes |
Theorem 2.82. (Stone's Theorem) Every metric space is paracompact. | Proof. The theorem was first proved by Arthur Stone Stone 48 in 1948. A much shorter proof can be found in [RudiM69]. | No |
Lemma 3.4. Let \( {X}_{1},\ldots ,{X}_{k} \) be topological spaces. The set \[ \mathcal{B} = \left\{ {{U}_{1} \times \cdots \times {U}_{k} \mid \text{ each }{U}_{i}\text{ is open in }{X}_{i}}\right\} \] has the basis property from page 102. | Proof (*). We need to verify that \( \mathcal{B} \) satisfies the conditions (B1) and (B2) formulated on page \( \left\lbrack {102}\right\rbrack \) (B1) We have \( {X}_{1} \times \cdots \times {X}_{k} \in \mathcal{B} \) . Thus given any \( \left( {{x}_{1},\ldots ,{x}_{k}}\right) \in {X}_{1} \times \cdots \times {X}_{k}... | Yes |
Lemma 3.5. Let \( {k}_{1},\ldots ,{k}_{m} \in {\mathbb{N}}_{0} \) . The map\n\n\[ \n{\mathbb{R}}^{{k}_{1}} \times \cdots \times {\mathbb{R}}^{{k}_{m}} \rightarrow {\mathbb{R}}^{{k}_{1} + \cdots + {k}_{m}} \n\]\n\n\[ \n\left( {\left( {{x}_{1},\ldots ,{x}_{{k}_{1}}}\right) ,\ldots ,\left( {{x}_{1},\ldots ,{x}_{{k}_{m}}}\... | Proof \( \left( *\right) \) . It is clear that the map is a bijection. So it remains to show that the map and its inverse are continuous. The continuity of both maps can be shown using Proposition 2.37 and Lemma 2.5. We leave it to the reader to fill in the details. | No |
Lemma 3.6. Let \( {X}_{1},\ldots ,{X}_{k} \) be topological spaces.\n\n(1) For each \( i \in \{ 1,\ldots, k\} \) the projection map\n\n\[ \n{p}_{i} : {X}_{1} \times \cdots \times {X}_{k} \rightarrow {X}_{i} \]\n\n\[ \n\left( {{x}_{1},\ldots ,{x}_{k}}\right) \mapsto {x}_{i} \]\n\nis continuous. | Proof.\n\n(1),(2) These two statements follow easily from the definition. | No |
Lemma 3.9. (*) Let \( {X}_{1},\ldots ,{X}_{k} \) be topological spaces.\n\n(1) If \( {A}_{i} \subset {X}_{i}, i = 1,\ldots, k \) are closed subsets, then \( {A}_{1} \times \cdots \times {A}_{k} \) is a closed subset of\n\n\( {X}_{1} \times \cdots \times {X}_{k} \) | Proof (*).\n\n(1) We have \n\nThus we see that \( \left( {{X}_{1} \times \cdots \times {X}_{k}}\right) \smallsetminus \left( {{A}_{1} \times \cdots \times {A}_{k}}\right) \) is the intersection of finitely many open ... | Yes |
Lemma 3.10. (*) Let \( X, Y \) and \( Z \) be topological spaces and let \( f : X \times Y \rightarrow Z \) be a map.\n\nSuppose we are given finitely many closed subsets \( {A}_{1},\ldots ,{A}_{m} \) of \( X \) with \( \mathop{\bigcup }\limits_{{i = 1}}^{m}{A}_{i} = X \) . If each map \( {A}_{i} \times Y \rightarrow X... | Proof (*). By Lemma 3.8 the identity\n\n\( \begin{matrix} {A}_{i}\text{ equipped with the } \\ \text{ subspace topology from }X \end{matrix} \times Y \rightarrow \begin{matrix} {A}_{i} \times Y\text{ equipped with the subspace } \\ \text{ topology from }X \times Y \end{matrix} \)\n\n is actually a homeomorphism. Togeth... | Yes |
Lemma 3.11. (*) Let \( {X}_{1},\ldots ,{X}_{k} \) be topological spaces and let \( {\mathcal{B}}_{1},\ldots ,{\mathcal{B}}_{k} \) be bases for these topological spaces. Then\n\n\[ \mathcal{C} \mathrel{\text{:=}} \left\{ {{B}_{1} \times \cdots \times {B}_{k} \mid \text{ for each }i \in \{ 1,\ldots, k\} \text{ we have }{... | Proof (*). The main idea is to prove the lemma using Lemma 2.27 (1). Therefore let \( W \subset {X}_{1} \times \cdots \times {X}_{k} \) be an open set and let \( \left( {{x}_{1},\ldots ,{x}_{k}}\right) \in W \) . By definition of the product topology there exist open sets \( {U}_{i} \subset {X}_{i}, i = 1,\ldots, k \) ... | Yes |
Lemma 3.13. We consider the map \( \Theta : {\mathbb{R}}^{2} \times {S}^{1} \rightarrow {\mathbb{R}}^{3} \)\n\n\[ \left( {\left( {x, y}\right) ,{e}^{\mathrm{i}\varphi }}\right) \mapsto \underset{\text{rotation around }z\text{-axis }}{\underbrace{\left( \begin{matrix} \cos \varphi & - \sin \varphi & 0 \\ \sin \varphi & ... | Proof (*). It follows easily from the example on page 163 that the map \( \Theta : {S}^{1} \times {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{3} \) is continuous. An elementary argument shows that the restriction of \( \Theta \) to \( {\bar{B}}^{2} \times {S}^{1} \) is injective. Since the solid torus \( {\bar{B}}^{2} \... | Yes |
Lemma 3.14. Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a family of topological spaces. We set\n\n\[ \n\left. \begin{array}{ll} \text{ (1) } & \text{ each }{U}_{i}\text{ is open in }{X}_{i}\text{ and } \\ \text{ (2) } & \text{ there exist only finitely many }i\text{ ’s with }{U}_{i} \neq {X}_{i} \end{array}\right... | Proof. Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a family of topological spaces.\n\n(0) The proof that \( \mathcal{B} \) has the basis property is basically identical to the proof of Lemma 3.4,\n\n(1) It follows immediately from the definitions that the projection maps \( {p}_{i} \) are continuous with respect ... | Yes |
(1) (a) If \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is a finite family of topological spaces, i.e. if \( I \) is finite, then the above product topology agrees with the product topology introduced on page 162. | Proof. All statements follow easily from the definitions and/or from Lemma 3.14. We leave it to the reader to provide the details. | No |
Proposition 3.16. Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a family of non-empty topological spaces.\n\n(1) The product \( \mathop{\prod }\limits_{{i \in I}}{X}_{i} \) is Hausdorff if and only if each \( {X}_{i} \) is Hausdorff. | The proof of all statements of Proposition 3.16 is basically the same as the proof of Proposition 3.12, except, for the \ | No |
Theorem 3.17. (Tychonoff’s Theorem \( {}^{52} \) ) The product of arbitrarily many compact topological spaces is again a compact topological space. | Sketch of Proof. We will not really make use of Tychonoff's Theorem thus we limit ourselves to a sketch of the proof for \( I = \mathbb{N} \) . Thus let \( {\left\{ {X}_{i}\right\} }_{i \in \mathbb{N}} \) be a family of compact topological spaces. Furthermore let \( {\left\{ {U}_{j}\right\} }_{j \in J} \) be a family o... | No |
Corollary 3.19. Let \( X \) be a metric space and let \( K \) and \( L \) be two non-empty subsets. We assume that one of the following holds:\n\n(i) Both \( K \) and \( L \) are compact, or\n\n(ii) \( X = {\mathbb{R}}^{n}, K \) is compact and \( L \) is a closed subset of \( X = {\mathbb{R}}^{n} \) .\n\nThe following ... | Proof.\n\n(1) First we suppose that we are in the setting (i). It follows immediately from our hypothesis and Proposition 3.12 that \( K \times L \) is compact. As we pointed out above, the metric \( d : K \times L \rightarrow {\mathbb{R}}_{ \geq 0} \) is continuous. The desired statement is now an immediate consequenc... | Yes |
Lemma 3.21. Let \( \sim \) be an equivalence relation on a topological space \( X \) . If we denote by \( p : X \rightarrow X/ \sim \) the canonical projection map from \( X \) onto the set of equivalence classes \( X/ \sim \), then the following statements hold:\n\n(1) The set\n\n\[ \mathcal{T} \mathrel{\text{:=}} \le... | (1) This statement follows immediately from Lemma 1.3 (5) and (6). | No |
Lemma 3.22. Let \( \sim \) be an equivalence relation on a topological space \( X \) and let \( f : X \rightarrow Y \) be a map with the property that \( f\left( x\right) = f\left( y\right) \) whenever \( x \sim y \) . Then the map\n\n\[ g : X/ \sim \rightarrow Y \]\n\n\[ \left\lbrack x\right\rbrack \mapsto f\left( x\r... | Proof (*). Let \( \sim \) be an equivalence relation on a set \( X \) and let \( f : X \rightarrow Y \) be a map with the property that \( f\left( x\right) = f\left( y\right) \) whenever \( x \sim y \) . Let \( a \) be an element of \( X/ \sim \) . By definition there exists an \( x \in X \) with \( a = \left\lbrack x\... | Yes |
Lemma 3.24. (*) Let \( X \) be a topological space and let \( A \subset X \) be a subset. Furthermore let \( Y \) be another topological space and let \( {y}_{0} \in Y \) . The map\n\n\[ \n\Phi : \left\{ \begin{matrix} \text{ set of continuous maps }f : X \rightarrow Y \\ \text{ such that }f\left( A\right) = \left\{ {y... | Proof (*). It follows from Lemma 3.22 that given a continuous map \( f : X \rightarrow Y \) with \( f\left( A\right) = \left\{ {y}_{0}\right\} \) the corresponding map \( \overline{f : }X/A \rightarrow Y \) is continuous. This shows that the map \( \Phi \) is well-defined. It is clear that \( \Phi \) is injective. Fina... | Yes |
(1) The map\n\n\[ \n\begin{aligned} \left( {\left\lbrack {0,1}\right\rbrack \times {S}^{n}}\right) /\left( {\{ 0\} \times {S}^{n}}\right) & \rightarrow {\bar{B}}^{n + 1} \\ \left\lbrack \left( {r, v}\right) \right\rbrack & \mapsto r \cdot v \end{aligned} \n\]\n\nis a homeomorphism. | (1) It follows from Lemma 3.22 that the given map is continuous. It is basically clear that the map is a bijection. The left-hand side is compact by Lemma 3.22 (4). It follows from Proposition 2.43 (3) that the map is a homeomorphism. | No |
Lemma 3.26. (*) Let \( X \) be a topological space that is Hausdorff and compact. If \( \sim \) is a closed relation, then \( X/ \sim \) is Hausdorff. | Proof (*). We denote by \( p : X \rightarrow X/ \sim \) the projection. We start out with the following claim.\n\nClaim. Given any closed subset \( C \) of \( X \) the projection \( p\left( C\right) \) is a closed subset of \( X/ \sim \) .\n\nLet \( C \) be a closed subset of \( X \) . We need to show that \( {p}^{-1}\... | Yes |
Lemma 3.27. (*) Let \( X \) be a topological space and let \( \sim \) be an equivalence relation on \( X \) . Furthermore let \( \mathcal{B} \) be a basis of the topology of \( X \) . If the projection \( p : X \rightarrow X/ \sim \) is an open map, then\n\n\[ p\left( \mathcal{B}\right) = \{ p\left( B\right) \mid B \in... | Proof (*). First note that the sets in \( p\left( \mathcal{B}\right) \) are indeed open subsets of \( X/ \sim \) since we assume that the projection is open. We use the criterion provided by Lemma 2.27 (1) to show that \( p\left( \mathcal{B}\right) \) is a basis for the topology of \( X/ \sim \) . Thus let \( U \subset... | Yes |
Lemma 3.29. (*) Let \( X \) be a topological space and let \( A \) be a closed subset.\n\n(1) For every open subset \( U \subset X \smallsetminus A \) the obvious map \( U \rightarrow X/A \) is an open embedding.\n\n(2) For every open neighborhood \( W \) of \( A \) the obvious map \( W \smallsetminus A \rightarrow W/A... | Proof (*). Let \( X \) be a topological space and let \( A \) be a closed subset. In the following we denote by \( p : X \rightarrow X/A \) the projection.\n\n(1) Let \( U \subset X \smallsetminus A \) be an open subset. We want to show that \( q : U \rightarrow X/A \) is an open embedding. It follows from \( U \subset... | Yes |
Let \( X \) be a topological space and let \( G \) be a group that acts on \( X \). (1) The projection map \( X \rightarrow X/G \) is continuous. (2) If \( G \) acts continuously on \( X \), then the projection map is also open. | Proof. The first statement follows immediately from Lemma 3.21 (3). We turn to the proof of the second statement. Thus let \( U \subset X \) be an open subset. We need to show that \( p\left( U\right) \) is open in \( X/G \) . Thus we need to show that \( {p}^{-1}\left( {p\left( U\right) }\right) \) is an open subset o... | Yes |
Lemma 3.31. The map\n\n\[ \n{\mathbb{R}}^{n}/{\mathbb{Z}}^{n} \rightarrow {\left( {S}^{1}\right) }^{n} \]\n\n\[ \n\left\lbrack \left( {{t}_{1},\ldots ,{t}_{n}}\right) \right\rbrack \mapsto \left( {{e}^{{2\pi }\mathrm{i}{t}_{1}},\ldots ,{e}^{{2\pi }\mathrm{i}{t}_{n}}}\right) \]\n\nis a homeomorphism. | Proof. We start out with the following observations:\n\n(1) The map is continuous by Lemma 3.22 and it is easy to see that it is a bijection.\n\n(2) The obvious map \( {\left\lbrack 0,1\right\rbrack }^{n} \rightarrow {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}/{\mathbb{Z}}^{n} \) is continuous and it is easily seen t... | Yes |
Lemma 3.32. Let \( n \in \mathbb{N} \). (1) The maps \[ \begin{aligned} \mathrm{O}\left( {n + 1}\right) \times {S}^{n} & \rightarrow {S}^{n} & & \\ \left( {A, v}\right) & \mapsto A \cdot v & & \text{ and } \end{aligned}\;\begin{aligned} \mathrm{O}\left( {n + 1}\right) \times {\bar{B}}^{n + 1} & \rightarrow {\bar{B}}^{n... | Proof. Statements (1) and (2) can be verified easily. | No |
Lemma 3.34. (*) Let \( G \) be a group that acts continuously and properly on a topological space. If \( X \) is Hausdorff, then given any two points a and \( b \) in \( X \) there exist open neighborhoods \( A \) of \( a \) and \( B \) of \( b \) with the following property: for every \( g \in G \) with \( {ga} \neq b... | Let \( a \) and \( b \) be two points in \( X \). Since \( G \) acts properly on \( X \) there exist open neighborhoods \( U \) of \( a \) and \( V \) of \( b \) such that \( \{ g \in G \mid {gU} \cap V \neq \varnothing \} \) is a finite set. We denote the elements of this finite set by \( {g}_{1},\ldots ,{g}_{r} \). W... | Yes |
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