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I'm having problems with the notation of this exercise:

Let $f:X\to Y$ be a map between topological spaces. Prove that are equivalent:

  1. $f$ is continuous.
  2. $f^{-1}(Int(B)) \subset Int(f^{-1}(B))$, for all $B\subset Y$.

I know that the elements of a topology are open subsets, but what about the elements of a topological space? That $B$ can be closed or it's always open?

Thanks and sorry if my question is too silly, but I don't get the difference.

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  • $\begingroup$ Here $B$ is not an element but a subset of a topological space. $\endgroup$ – drhab Nov 18 '17 at 9:04
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The set $B$ can by any subset of $Y$. It can be closed, it can be open and it can be neither closed nor open.

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    $\begingroup$ Also it can be both closed and open. $\endgroup$ – Adayah Nov 18 '17 at 8:03
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The term topology refers to a subset $B$ of $P(X)$ satisfying some conditions, whereas the term topological space refers to the 2-tuple $(X,B)$.

So, an element of a topological space is just an element of $X$.

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$B$ is an arbitrary subset of $Y$. For any subset of a topological space we have the interior of $B$ ($\operatorname{int}(B)$) which is the largest open subset of $B$ (where an open set is a subset of $Y$ that happens to be in the topology of $Y$; that's all a topology is: it's exactly the collection of all subsets of $Y$ that are called open, and to make this meaningful (non-trivial), this collection has to be obey some axioms that you probably know). So $B$ is any set but $\operatorname{int}(B)$ is an open set. Think of the topology on $Y$ as the "nice" or "special" subsets in some way.

Now assume $f$ is continuous (1) and $B$ is any subset of $Y$ then we know that $f^{-1}[ \operatorname{int}(B)]$ is also open in $X$, as the inverse image of an open set.

And $f^{-1}[B]$ is just a well-defined subset of $X$ (all $x \in X$ such that $f(x) \in B$), and so it too has an interior (but in $X$ of course).

Because $\operatorname{int}(B) \subset B$ by definition, we also have $f^{-1}[\operatorname{int}(B)] \subset f^{-1}[B]$ (if an $x$ maps into $\operatorname{int}(B)$ it surely maps into $B$ as well).

And as $\operatorname{int}(f^{-1}[B])$ is the largest open subset of $f^{-1}[B]$ and $f^{-1}[\operatorname{int}(B)]$ is one such open subset of $f^{-1}[B]$ we know that $$f^{-1}[\operatorname{int}(B)] \subset \operatorname{int}(f^{-1}[B])]$$ when $f$ is continuous.

This shows that 1 implies 2.

For the reverse use that $O$ is open iff $\operatorname{int}(O) = O$, and assuming the formula $2$ (which holds for all subsets of $Y$) prove that the inverse image of an open set of $Y$ is open in $X$.

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