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Chapter 10: Problem 6

Show that a map \(f \cdot X \rightarrow Y\) between two topological spaces \(X\) and \(Y\) is contintous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for all sets \(U \subseteq X\) Show that \(f\) is a lwomeomorplism only if \(f(\bar{U})=\overline{f(U)}\) for all sets \(U \subseteq X\)

Short Answer

Expert verified
So the function \(f: X \rightarrow Y\) is continuous if and only if for any subset \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). And \(f\) is a homeomorphism if and only if for any subset \(U \subseteq X\), \(f(\bar{U})= \overline{f(U)}\). Therefore, it's been demonstrated that these conditions are both necessary and sufficient for continuity and homeomorphism.

Step by step solution

01

Prove \(f\) is Continuous

Assume that for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Let \(V\) be an arbitrary closed subset of \(Y\). We must prove that \(f^{-1}(V)\) is also closed in \(X\). The complement \(X - f^{-1}(V)\) is an arbitrary subset of \(X\). Using the assumption, we have \(f(\overline{X - f^{-1}(V)}) \subseteq \overline{f(X - f^{-1}(V))}\). As \(f^{-1}(V)\) equals \(X - \overline{X - f^{-1}(V)}\), it follows that \(f^{-1}(V)\) is closed.
02

Show the Bi-conditional for the Continuity of \(f\)

Now we have to show the reverse, i.e., if \(f\) is continuous then for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Assume \(f\) is continuous, and let \(U\) be an arbitrary subset of \(X\). We must prove that \(f(\bar{U}) \subseteq \overline{f(U)}\). Since \(f\) is continuous, \(f(\bar{U})\) is closed in \(Y\), which means \(f(\bar{U}) \subseteq \overline{f(U)}\). Therefore, \(f\) is continuous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for every \(U \subseteq X\).
03

Prove \(f\) is a Homeomorphism

Again, let's assume that for every set \(U \subseteq X\), \(f(\bar{U}) = \overline{f(U)}\). From Step 2, it already follows that \(f\) is continuous. To prove that \(f\) is a homeomorphism, we must show that its inverse \(f^{-1}\) is also continuous. By applying the condition to \(f^{-1}\), we get that \(f^{-1}(\bar{V}) = \overline{f^{-1}(V)}\) for every \(V \subset Y\), thus proving that \(f^{-1}\) is continuous.
04

Show the Bi-conditional for Homeomorphism of \(f\)

For the reverse, assume that \(f\) is a homeomorphism. As a result, both \(f\) and \(f^{-1}\) are continuous (From Steps 1 and 3). Now we must prove that \(f(\bar{U}) = \overline{f(U)}\) for every \(U \subset X\). We already know from Step 2 that \(f(\bar{U}) \subseteq \overline{f(U)}\). For the reverse inclusion, denote \(V = f(U)\). As \(f^{-1}\) is continuous, we have \(f^{-1}(\overline{V}) \subseteq \overline{f^{-1}(V)} = \overline{U}\). As \(f(\overline{U}) \supseteq f(f^{-1}(\overline{V})) = \overline{V}\), we have proved the required.

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Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\).

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