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I am used to the definition of continuity used in real analysis, based on the idea that if $f$ is continuous at $a$, you can get arbitrarily close to $f(a)$ by going arbitrarily close to $a$.

However, the topological concept of continuity of $f: M \to N$ is the condition that for all $S \subset N$, the preimage of $S$ in $M$ must be open subsets.

What does this have to do with the intuitive concept of "continuity"? I cannot even picture how this relates to continuity in Euclidian space, let alone how it would relate to continuity in more abstract topological spaces.

Arthur
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user56834
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    The first step is to write up the corresponding definition of continuity at a point, rather than the one for the entire function. This will not immediately make it clear what is going on, but it is a better place to start. – Tobias Kildetoft Jan 31 '17 at 09:01
  • Look at a point $x_0$ of discontinuity of a non-complicated function $f:\Bbb R\to \Bbb R$ (e.g. a function consisting of two linear parts that do not meet up), and focus on the preimages of the open sets $(f(x_0)-\epsilon,f(x_0)+\epsilon)$ for positive, small values of $\epsilon$. That should give you an idea that at least the definitions try to catch the same kind of behaviour. – Arthur Jan 31 '17 at 09:01
  • You can refer this, this and this. – Error 404 Jan 31 '17 at 09:08

3 Answers3

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Take the case of a metric space, this is the same as what you are used to. So say $f:X\rightarrow Y$ is a map between metric spaces. I believe you will agree about the following definition of continuity at $a\in X$,

$$\forall \epsilon \ \ \exists \delta \ \ \forall x\in X\ \ \ [d_X(x,a)<\delta \rightarrow d_Y(f(a),f(x))]$$ Now define $$B_{\epsilon}(x)=\{y\in X |d(x,y)<\epsilon\}$$

by definition these sets are a base for the topology on a metric space. Lets now rewrite the above definition in this notation $$\forall \epsilon \ \ \exists \delta \ \ \forall x\in X\ \ \ [x\in B_{\delta}(a) \rightarrow f(x)\in B_{\epsilon}(f(a))]$$

Now lets phrase it in such a way that the $\forall x\in X$ does not appear

$$\forall \epsilon \ \ \exists \delta \ \ \ \ [ f(B_{\delta}(a)) \subseteq B_{\epsilon}(f(a))]$$

which is equivalent to $$\forall \epsilon \ \ \exists \delta \ \ \ \ [ B_{\delta}(a) \subseteq f^{-1} (B_{\epsilon}(f(a)))]$$

This says that the inverse of an open set about $f(a)$ contains an open set about $a$.

We could write it as

$$\forall \text{$U$ open neighbourhood of $f(a)$} \exists \text{$V$ open neighbourhood of $a$}[V\subseteq f^{-1}(U)]$$

If we now assume that $f$ is continuous at all $a$ this means that

$$\forall \text{$U$ open neighbourhood of $f(a)$} [f^{-1}(U)\text{is open}]$$ and again since this holds for all $a$, $$\forall \text{$U$ open set in $Y$} [f^{-1}(U)\text{is open}]$$

Thus we have the definition of continuity in a topological space.

Rene Schipperus
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The intuition for topological continuity is really similar to real analysis. The difference is that the set of all sequences is not "big" enough in topological context. Metric spaces are first countable while in general topological spaces don't have to be. But if you replace the word "sequence" with the word "net" (see https://en.wikipedia.org/wiki/Net_(mathematics) ) then you get the same intuition, i.e.

A function is continous if and only if it maps convergent nets to convergent nets.

Note that a sequence is a special case of net. Virtually any topological concept can be rephrased in terms of nets. But I guess mathematicians don't find nets easy to work with.

freakish
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Let $X,Y$ be topological spaces. A function $f:X\to Y$ is continuous iff the following holds:

for any point $x_0\in X$ and any neighborhood $V$ of $f(x_0),$ there is a neighborhood $U$ of $x_0$ such that $x\in U\implies f(x)\in V.$

The condition "every open set $S\subseteq Y$ has an open preimage in $X$" is just a slick way of saying the same thing.

bof
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