Continuous function

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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