Let $(X, \mathcal{T})$, $(Y, \mathcal{U})$ be topological spaces, and let $f: X \rightarrow Y$ be a bijection. Then $f$ is open iff $f$ is closed

Question

Here's my proof of the above statement.

Since $f$ is open, it maps open sets in $\mathcal{T}$ to open sets in $\mathcal{U}$. Let $U \in \mathcal{T}$ be a closed set. So, $X \setminus U$ is an open set. By definition, it maps $X \setminus U$ to $Y \setminus V$ where $V$ is a closed set in $\mathcal{U}$. Thus it maps the complement of every closed set in $\mathcal{T}$ to the complement of some other closed set in $\mathcal{U}$. So the function induces a mapping between closed sets via their complements. Thus $f$ is closed. Similarly, one can prove the other implication as well.

Is my proof correct?

Answer 1

I would not say "by definition it maps $X\setminus U$ to $Y\setminus V$" - it follows from no definition. It follows from the fact that $f$ is a bijection, and you may want to elaborate how exactly it follows.

Also, you started with a closed subset $U$ so you should proceed with $U$ until the end:

• $U$ is closed,
• Therefore, $X\setminus U$ is open,
• Therefore, $Y\setminus V$ is open (as $f$ is an open map and you've just claimed that it maps $X\setminus U$ to $Y\setminus V$),
• Therefore, $V$ is closed,

... and there the proof is done: an arbitrary closed set $U$ is mapped to a closed set $V$, so $f$ is a closed map. Short and up to the point.

Instead you go with:

Thus it maps the complement of every closed set in $\mathcal T$ to the complement of some other closed set in $\mathcal U$. So the function induces a mapping between closed sets via their complements."

This is all true but is wordy and does not really develop the argument. I am not even sure what those two sentences claim that hasn't been said previously, or how they are connected. It doesn't seem that one follows from the other. It is like you are trying to describe in other words what you have previously already done - that is, without fully having it done.

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By the way, if you know the definition of the term continuous function, and if you know the equivalence:

• Function $\phi:(Y,\mathcal U)\to(X, \mathcal T)$ is continuous
• The preimage $\phi^{-1}(U)$ of every open subset $U\subset X$ is open
• The preimage $\phi^{-1}(U)$ of every closed subset $U\subset X$ is closed

and if you know that, if $f$ is a bijection and $\phi=f^{-1}$ then the direct image $f(U)$ is the same as the preimage $\phi^{-1}(U)$, then you can have a proof like this:

• $f$ is open, i.e. direct image of every open set is open
• $\phi$ is continuous, i.e. preimage of every open set is open
• $\phi$ is continuous, i.e. preimage of every closed set is closed
• $f$ is closed, i.e. direct image of every closed set is closed

are all equivalent.