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This is an extremely common theorem, I have a function $f$ that maps $f:(X,\mathscr{S})\to(Y,\mathscr{T})$. I want to show that $f$ is continuous if and only if for all $V\in \mathscr{T}$, $f^{-1}(V)\in \mathscr{S}.$

I know exactly how to do this using distant function, but now I want to prove it using the definition of continuity in a topological space.

I did the forward direction by assuming that $f$ is continuous at $x\in X$, then $\forall f(x)\in V\in \mathscr{T}$, $\exists \,U$ such that $f(U)\subset V$.

Then because $f$ is continuous, I took $f^{-1}$ of both sides (actually this is the step I questioned most), then I get $U\subset f^{-1}(V)$. But then what do I do next? Can I conlude that this implies that $f^{-1}(V)\in \mathscr{S}$?

idm
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Akaichan
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    Isn't the first paragraph the definition of continuity? – Arthur Aug 26 '14 at 14:20
  • Echoing what Arthur said, if that isn't your definition of continuous, what is? – Hayden Aug 26 '14 at 14:22
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    that's what I was trying to prove, the definition is the one that says $f(U)\subset V$ – Akaichan Aug 26 '14 at 14:22
  • This is the definition for continuity I am using: $f$ is continuous at $x\in X$, then $\forall f(x)\in V\in \mathscr{T}$, $\exists ,U$ such that $f(U)\subset V$. The first paragraph is a theorem similar to a theorem in real analysis, I just want to do this using the definition from topology – Akaichan Aug 26 '14 at 14:24
  • @Akaichan Just so you know, what you're trying to prove is the standard definition of continuous. –  Aug 26 '14 at 14:28
  • @MikeMiller: But then why is it a theorem in real analysis/ metric spaces? Isn't the definition for continuity what I stated above? – Akaichan Aug 26 '14 at 14:31
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    The definition you are using (according to your comment) is a definition of continuous at $x\in X$. So not a definition of continuous. BTW the set $U$ you mention must contain $x$. – drhab Aug 26 '14 at 14:37
  • Oh, I understand what you are trying to say now – Akaichan Aug 26 '14 at 14:42

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Maybe this is helpful.

Let it be that $f:X\rightarrow Y$ is continuous at each element $x\in X$.

If $V$ is open in $Y$ then for each $x\in X$ with $f(x)\in V$ there is an open set $U_x\subset X$ with $x\in U_x$ and $f(U_x)\subset V$ or equivalently $U_x\subset f^{-1}(V)$.

As union of open sets $\bigcup_{x\in f^{-1}\left(V\right)}U_{x}$ is open in $X$ and $f^{-1}(V)=\bigcup_{x\in f^{-1}\left(V\right)}U_{x}$ .

Proved is now that a function that is continous at each element of $x\in X$ is continuous.

Also the converse is true:

If $f:X\rightarrow Y$ is continuous and $x\in X$ with $f\left(x\right)\in V$ and $V$ open in $Y$ then $U:=f^{-1}\left(V\right)$ is an open set with $x\in U$ and $f\left(U\right)\subset V$. This shows that $f$ is continuous at $x$, and works for every $x\in X$.

drhab
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