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I am trying to learn how to prove that the preimage of an open set is open in general topology. Here is an example that I am not really satisfied with

Proposition 3.9 (Book: Essential Topology, Crossley): If $S$ has the discrete topology and $T$ is any topological space, then any function $f:S\to T$ is continuous.

Proof: If $S$ has the discrete topology, then every subset of $S$ is open, so in particular, every preimage of an open set must be open. Thus $f$ is continuous.

Here is my proof: Let $(X,\mathfrak{T}_{X})$ and $(Y,\mathfrak{T}_{Y})$ be topological spaces, where $\mathfrak{T}_{X}$ is the discrete topology, while $\mathfrak{T}_{Y}$ is any topology on $Y$. We wish to show that $U\in \mathfrak{T}_{Y}\implies f^{-1}(U)\in \mathfrak{T}_{X}$.

Assume that $U$ is open in $Y$. Pick $a\in f^{-1}(U)$. Then $f(a)\in U$. Since $U$ is open, there exists neighbourhood $V_{a}$ of $a$ such that $f(V_{a})\subseteq U$. Then $V_{a}\subseteq f^{-1}(U)$. My thought is now to make $f^{-1}(U)$ "completely" open that contains a lot of neighbourhoods. Letting $V=\bigcup_{x\in f^{-1}(U)}V_{x}$, so $V_{a}\subseteq V=f^{-1}(U)$. Hence $f^{-1}(U)$ is open in $X$. $\square$

How is the proof so far? Any improvings or ideas are much appreciated.

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    $\begingroup$ "since $U$ is open there exists a neighborhood $V_a$ of $a$ such that $f(V_a)\subseteq U$". This is not correct in general. Also you are making things unnecessary difficult. $f^{-1}(U)$ is open simply because any subset of $X$ is open with respect to the discrete topology. $\endgroup$ – drhab Dec 15 '15 at 19:16
  • $\begingroup$ @drhab Ok, you are saying that the preimage of every open subset of Y is open in X, because every subsets of a topology on X is open. I don't need to worry about such preimage itself is contained in that topology, right? $\endgroup$ – UnknownW Dec 15 '15 at 19:32
  • $\begingroup$ The preimages of open subsets of $Y$ are automatically subsets of $X$, hence are open in $X$ because every subset of $X$ is open. This allows you to conclude that every function $f:X\to Y$ is continuous. $\endgroup$ – drhab Dec 15 '15 at 19:47
  • $\begingroup$ @drhab Oh I see. Nice. I now understand better. Thank you so much! $\endgroup$ – UnknownW Dec 15 '15 at 19:54
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Since $U$ is open, there exists a neighbourhood $V_a$ of $a$ such that $f(V_a) \subseteq U$.

This is not necessarily the case. Since $U$ is open, you know a neighborhood of $f(a)$ is contained in $U$. That is not the same thing as saying that the image of a neighborhood of $a$ is contained in $U$.

The real problem, though, is that you have not used anywhere in your proof that the topology on $X$ is the discrete topology. If your proof were correct, it would imply every map between topological spaces is continuous, which is not the case. The proof really is as simple as the one in the textbook. You wish to show that given an open subset $U$ of $Y$, $f^{-1}(U)$ is an open subset of $X$. Well every subset of $X$ is open, so $f^{-1}(U)$ is an open subset of $X$.

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  • $\begingroup$ I know well that every sets of topology on X are open. The reason why I prove this way was because that I was worrying if preimage itself is contained the topology in X. But it seems that I don't need to, right? You are right that I didn't mention about this topology. $\endgroup$ – UnknownW Dec 15 '15 at 19:50
  • $\begingroup$ It looks like you've understood this from the comments above. The preimage $f^{-1}(U)$ is, by definition, a subset of $X$: $f^{-1}(U)=\{x \in X ~:~ f(x) \in U\}$. As $f^{-1}(U)$ is a subset of $X$, it is in the discrete topology on $X$. $\endgroup$ – kccu Dec 15 '15 at 20:38

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