Continuous function

Submitted by Structure on Sat, 11/17/2007 - 20:23.

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces and $f:X\rightarrow Y$ a function.The function f is said continuous in $x\in X$ if and only if for all V a neighborhood of f(x) we have $f^{-1}(V)$ is a neighborhood of x. f is continuous on X if and only if f is continuous in any $x\in X$ .
Theorem
Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces and $f:X\rightarrow Y$ a function. The following statement are equivalent
(i) f is continuous on X;
(ii) $\forall G\in\mathcal{T}_Y,$ we have $f^{-1}(G)\in \mathcal{T}_X$ ;
(iii) $\forall F$ closed in $Y$ we have $f^{-1}(F)$ is closed in $X$ ;
(iv) $\forall T \subseteq Y,$ we have $\overline{f^{-1}(T)}\subseteq f^{-1}(\overline{T})$ ;
(v) $\forall B\subseteq Y,$ we have $f^{-1}(\r{B})\subseteq \r{f}^{-1}(B)$

Proof.
$(i)\Rightarrow (ii)$ Let $\forall G\in\mathcal{T}_Y,$ . $f^{-1}(G)\in \mathcal{T}_X$ means that $f^{-1}(G)$ is open in $X$ . But a set is open if and only if the set is a neighborhood for any of its points.
So, let $x\in f^{-1}(G)$ then $f(x)\in G$ but $G$ is open in $Y$ so $G\in \mathcal{V}_{f(x)}$ and as f is continuous in x we have $f^{-1}(G)\in \mathcal{V}_x$ and so $f^{-1}(G)$ is a neighborhood for any of its points, so it is open.
$(ii)\Rightarrow (iii)$ Let F be closed in Y. Then $C_YF$ is open in Y and by (ii) $f^{-1}(C_YF)=C_Xf^{-1}(F)$ is open in X, so $f^{-1}(F)$ is closed in X.
$(iii)\Rightarrow (iv)$ Let be $T\in Y$ . As $T\subseteq \overline{T}$ we have $f^{-1}(T)\subseteq f^{-1}(\overline{T})$ . But from (iii) as $\overline{T}$ is closed in $Y$ , $f^{-1}(\overline{T})$ is closed in $X$ and if a set $f^{-1}(T)$ is included in a closed set, its closure is equally included in the same closed set, so $\overline{f^{-1}(T)}\subseteq f^{-1}(\overline{T})$ ;
$(iv)\Rightarrow (v)$ Let $B\subseteq Y$ , then
$C_{X}f^{-1}(\r{B})=f^{-1}(C_{X}\r{B})= f^{-1}(\overline{C_{Y}B)}\supseteq \overline{f^{-1}(C_{X}B)}=\overline{C_{Y}f^{-1}(B)}$ = $C_{X}\r{f}^{-1}(B)$ or $f^{-1}(\r{B})\subseteq \r{f}^{-1}(B)$
$(v)\Rightarrow(i)$ Let $x\in X$ and let $V\in \mathcal{V}_{f(x)}$ , then
$f(x)\in \r{V}$ or $x\in f^{-1}(\r{V})\subseteq\r{f}^{-1}(V)$ and so $f^{-1}(V)\in \mathcal{V}_x$ so f is continuous in an arbitrary $x\in X$ so f is continuous on $X$ .

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Continuous function