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Continuity of function $f:Xto Y$ from topological space $X$ to $Y$ is defined by saying that for any open set $U_Y$, $f^{-1}(U_Y)$ is also an open set.

Intuitively, I find this weird. If we interpret “open set” informally as “a set whose elements are nearby eachother” (indeed it is a set which is a neighbourhood of all its elements), then it makes intuitive sense to say that a continuous function $f$ is a function that does not “rip elements away from its neighbours”, i.e. if you input an open set $U_X$ (a set whose elements are “nearby eachother”), then this should not produce a set where some elements are “not nearby eachother”, i.e. it should produce an open set.

So is there an intuitive explanation at this level of abstraction (i.e. without reference to metric spaces for example) of why we don’t define continuity as “for any open set $U_X$, $f(U_X)$ is an open set”?

The problem with your intuition is that an "open set" is not "a set whose elements are nearby each other". For example, considering the real numbers with the standard topology, the set $(0, infty)$ contains elements arbitrarily far away from each other, while ${0}$ contains elements extremely nearby each other.

A better intuition is: an open set $X$ is a set such that if $x in X$, then all points that are close to $x$ are also in $X$. This shows why the "forward definition" doesn't work: just because you are taking all points close to some $x$, does not mean that you should map onto all points close to $f(x)$ -- it just means that you should hit only points close to $f(x)$. But what does hitting only points close to $f(x)$ mean? It means that if you take all points $U$ close to $f(x)$, then $f^{-1}(U)$ should include all points close to $x$.

If you try to make the ideas in the previous paragraph precise and formal, you end up with the ordinary definition of continuity.

Edit: From the comments:

I still find your intuition hard to reconcile with the idea of “a discontinuous function rips points apart”.

Let us look at the converse, and take $f$ discontinuous. Informally, this means that there are $x$, $y$, which are close together, such that $f(x)$ and $f(y)$ are not close together. (Of course, to make this formal, you need to take at least one of $x$ or $y$ to be a sequence or even a net, etc.)

My intuition of an open set says: let $X$ be an open set, then $x in X$ if and only if $y in X$. Now let's see if the "forward definition of continuous" lets us prove that $f$ is discontinuous. Let's take any open set $X$. If $x notin X$ then also $y notin X$, and this doesn't seem to go anywhere. So let's look at open $X$ with $x, y in X$. Then $f(X)$ is also open, and therefore $f(x), f(y) in f(X)$ -- but this is precisely not what we wanted to prove.

Now let's apply my intuition to the "backward definition of continuous". Because $f(x), f(y)$ are far apart, there is an open set containing $f(x)$ but not $f(y)$. Let's call it $Y$. Then we have $x in f^{-1}(Y)$, but $y notin f^{-1}(Y)$. Thus $f^{-1}(Y)$ is not an open set, and $f$ is discontinuous.

To say that “$mathcal U=f^{-1}(mathcal O)$ is not open” for the open set $mathcal O$ means that the complement of $mathcal U$ approaches a point $xinmathcal U$.

But since $mathcal O$ is open, the images of points outside of $mathcal U$ approaching $x$ cannot approach the image of $x$. Thus $mathcal O$ witnesses that $x$ has been torn from $mathcal U^c$ by $f$.

The definition successfully carries the intuition to mention.... but perhaps its ”contravariantness” is tripping up your acceptance of the intuition.