- #1
ice109
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i've texed up three proofs in from elementary topology. can someone please check them?
actually i'll just retype them here for convenience
8.2.5
Let f: X_{\tau} \rightarrow Y_{\nu} be continuous and injective. Also let Y_{\nu} be Hausdorff.
Prove : X_{\tau} is Hausdorff.
Proof : Pick y_1 and y_2 in f(X). By injectivity of f there exist x_1 =
f^{-1}(y_1) and x_2 = f^{-1}(y_2) such that they are both unique. f(X) is Hausdorff so by Theorem 1 there exist disjoint open neighborhoods U and V of y_1 and y_2 respectively. Then f^{-1}(U \bigcap V) = f^{-1}(U) \bigcap f^{-1}(V) = \emptyset and by continuity f^{-1}(U) and f^{-1}(V) are open. Finally by definition of f^{-1} : x_1 \in f^{-1}(U) and x_2 \in f^{-1}(V) which, as stated previously, are two open disjoint sets in X_{\tau}. Hence X_{\tau} is Hausdorff.\\
Comments:
Injectivity is necessary. Take for example X = \{a,b\} and Y = \{a\} and to be f: X_{\mathcal{I}} \rightarrow Y_{\mathcal{I}}. Explicitly f(\{a,b\}) =\{a}\}. f is continuous, Y_{\mathcal{I}} is obviously Hausdorff and X_{\mathcal{I}} is obviously not.\\
This does not prove that Hausdorff is a strong topological property because we have proven a stronger converse. To prove that Hausdorff is a strong topological property we would have to have proven that f: X_{\tau} \rightarrow Y_{\nu} continuous, not necessarily injective, and X_{\tau} Hausdorff implies f(X) Hausdorff.
8.2.7
Let X_{\tau} be T_1 and A \subseteq T and x \in A'.\\
Prove : Any neighborhood of x intersects \textit{A} in infinitely many points.
Proof : Assume that there exists a neighborhood of x, in X_{\tau} that intersects \textit{A} in only finitely many points to derive a contradiction. Let N_x be such a neighborhood. N_x is T_1 by Theorem 1. Hence we can separate x from all points in N_x by other neighborhoods. Since there are finitely many points in N_x there are finitely many such neighborhoods. Let \textit{N} be the intersection of those neighborhoods. N -\{x\} therefore is itself a non-trivial open neighborhood of \textit{x} which does not intersect \textit{A}. This contradicts that x is a limit point of \textit{A} and therefore any neighborhood of x intersects \textit{A} in infinitely many points.
\textbf{8.3.4}\\
Let A \subseteq X_{\tau} and X_{\tau} be regular.\\
Prove : A is regular.
Proof : Pick A_1 \subseteq A, A_1 closed in the subspace topology, and x \in A - A_1. Then A_1 = B for some B \in \tau and we can find two open sets N_B and N_x by the the regularity of X_{\tau} which are disjoint. N_B \bigcap A and N_x \bigcap A are two disjoint sets in A which contain A_1 and x respectively. Therefore A is regular.
Theorem 1 : A subspace of a T_i space for i \leq 2 is T_i.
actually i'll just retype them here for convenience
8.2.5
Let f: X_{\tau} \rightarrow Y_{\nu} be continuous and injective. Also let Y_{\nu} be Hausdorff.
Prove : X_{\tau} is Hausdorff.
Proof : Pick y_1 and y_2 in f(X). By injectivity of f there exist x_1 =
f^{-1}(y_1) and x_2 = f^{-1}(y_2) such that they are both unique. f(X) is Hausdorff so by Theorem 1 there exist disjoint open neighborhoods U and V of y_1 and y_2 respectively. Then f^{-1}(U \bigcap V) = f^{-1}(U) \bigcap f^{-1}(V) = \emptyset and by continuity f^{-1}(U) and f^{-1}(V) are open. Finally by definition of f^{-1} : x_1 \in f^{-1}(U) and x_2 \in f^{-1}(V) which, as stated previously, are two open disjoint sets in X_{\tau}. Hence X_{\tau} is Hausdorff.\\
Comments:
Injectivity is necessary. Take for example X = \{a,b\} and Y = \{a\} and to be f: X_{\mathcal{I}} \rightarrow Y_{\mathcal{I}}. Explicitly f(\{a,b\}) =\{a}\}. f is continuous, Y_{\mathcal{I}} is obviously Hausdorff and X_{\mathcal{I}} is obviously not.\\
This does not prove that Hausdorff is a strong topological property because we have proven a stronger converse. To prove that Hausdorff is a strong topological property we would have to have proven that f: X_{\tau} \rightarrow Y_{\nu} continuous, not necessarily injective, and X_{\tau} Hausdorff implies f(X) Hausdorff.
8.2.7
Let X_{\tau} be T_1 and A \subseteq T and x \in A'.\\
Prove : Any neighborhood of x intersects \textit{A} in infinitely many points.
Proof : Assume that there exists a neighborhood of x, in X_{\tau} that intersects \textit{A} in only finitely many points to derive a contradiction. Let N_x be such a neighborhood. N_x is T_1 by Theorem 1. Hence we can separate x from all points in N_x by other neighborhoods. Since there are finitely many points in N_x there are finitely many such neighborhoods. Let \textit{N} be the intersection of those neighborhoods. N -\{x\} therefore is itself a non-trivial open neighborhood of \textit{x} which does not intersect \textit{A}. This contradicts that x is a limit point of \textit{A} and therefore any neighborhood of x intersects \textit{A} in infinitely many points.
\textbf{8.3.4}\\
Let A \subseteq X_{\tau} and X_{\tau} be regular.\\
Prove : A is regular.
Proof : Pick A_1 \subseteq A, A_1 closed in the subspace topology, and x \in A - A_1. Then A_1 = B for some B \in \tau and we can find two open sets N_B and N_x by the the regularity of X_{\tau} which are disjoint. N_B \bigcap A and N_x \bigcap A are two disjoint sets in A which contain A_1 and x respectively. Therefore A is regular.
Theorem 1 : A subspace of a T_i space for i \leq 2 is T_i.
