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- Feb 5, 2012

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Here's a question I solved but wants to check my proof. Let me know if you see any problems with my logic.

Problem:

Problem:

Let \(f:\,(X,\,d_x)\rightarrow (Y,\,d_y)\) be a map. Prove that \(f\) is continuous if and only if for each \(F\subset Y\) closed, then \(f^{-1}(F)\) is also closed.

**My Solution:**

\(F\subset Y\) is closed. Then for all \(\epsilon>0\) we have, \(B_{\epsilon}(f(x))\cap (F\,\backslash \,f(x))\neq \emptyset\) where \(f(x)\in F'\subseteq F\).

Hence there exist \(f(y)\in F\) where \(y\neq x\) such that,

\[d_{y}(f(x),\, f(y))<\epsilon\]

Suppose that \(f\) is continuous. Then for all \(\epsilon>0\) there exist \(\delta>0\) such that,

\[d_{y}(f(x),\,f(y))<\epsilon\mbox{ whenever }d_{x}(x,\,y)<\delta\]

Hence \(y\in B_{\delta}(x)\) and therefore \(B_{\delta}(x)\cap(f^{-1}(F)\,\backslash \,\{x\})\neq \emptyset\). Therefore, \(f^{-1}(F)\) is closed.

Conversely, suppose \(f^{-1}(F)\) is also closed (in addition to \(F\subset Y\) being closed). Then, for each \(x\in (f^{-1}(F))'\subseteq f^{-1}(F)\) we have \(B_{\delta}(x)\cap (f^{-1}(F)\,\backslash \,\{x\})\neq \emptyset\) for all \(\delta>0\). Hence there exist \(y\in f^{-1}(F)\) such that, \(d_{x}(x,\,y)<\delta\).

Since \(F\subset Y\) closed we know that,

\[d_{y}(f(x),\,f(y))<\epsilon\]

Hence \(d_{y}(f(x),\,f(y))<\epsilon\mbox{ whenever }d_{x}(x,\,y)<\delta\) and therefore, \(f\) is continuous.