# [SOLVED]Continuous Map and Closed Sets

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

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

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.