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On page 27 of Lee's Introduction to Topological Manifolds, he writes

In metric spaces, one usually first defines what it means to be continuous at a point...in topological spaces, continuity at a point is not such a useful concept.

Why not? We can define continuity at a point $x_0$ by the requirement that for all $A \in \mathcal{P}(X), \, x_0\in \overline{A}\implies f(x_0)\in \overline{f(A)}.$

I suppose it's not easy to say why something is not useful, but if anyone has some insight, I'd be glad to hear it.

Eric Auld
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  • He means that in a metric space a neighborhood of a point is well understood and can be described explicity, which is not the case for a general topological space. – palio Feb 15 '14 at 15:34
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    Of course continuity at a point is always not as useful as global continuity, but I see no reason why it would make a difference whether the space is metric or not. – Daniel Fischer Feb 15 '14 at 15:46
  • in Topology the big picture is more important,suppese you have two indentical spheres,delete one point from first one above and secand one below you have still two identical shape.You can think deleted points as discontinuity point, as you see it is not important at which point it has discontiniouty instead, it is important to how it affects the space in total. – mesel Feb 15 '14 at 16:42

2 Answers2

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One possible characterization of "continuity of $f:X\to Y$ at a point $x\in X$" for $X,Y$ arbitrary topological spaces is that $$ \tag{1} \text{for every net $x_\alpha \to x$, we have $f(x_{\alpha}) \to f(x)$.}$$

Suppose we define a neighborhood of $x$ as a set containing an open set containing $x$. I claim that an equivalent condition to (1) is

$$\tag{1'} \text{for every neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$}.$$

I claim further that (1) is equivalent to

$$\tag{1''}\text{for every open set $V$ containing $f(x)$, $f^{-1}(V)$ is open}.$$

Therefore, equivalent to continuity is that (1') holds at all points $x$.

Eric Auld
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    That last one isn't right. Let $X = {0,1,2}$ with topology $\tau_X = {\varnothing, {0}, {1,2},X}$, and $Y = {a,b,c}$ with the discrete topology. The map $0 \mapsto a,, 1 \mapsto b,, 2 \mapsto c$ is continuous at $0$, but the preimage of the open set ${a,b}$ containing $a$ is ${0,1}$, which is not open. – Daniel Fischer Jan 22 '17 at 12:57
  • But it seems some author use the preimage definition. I guess a continuous function from any space to discrete space should be locally constant because $f^{-1}({a})$ is constant for all domain elements such that $f(\star)=x$. @DanielFischer Is your mapping is locally constant? – WhyMeasureTheory Dec 23 '21 at 09:03
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In a metric space we begin with the $\epsilon$-$\delta$ definition of continuity, which is then globalized by the requirement that the $\epsilon$-$\delta$ holds at every point. This is a natural starting point because from the quantitative $\epsilon$-$\delta$ relation we are led to several related important concepts:

  • uniform continuity
  • Hölder continuity
  • Lipschitz continuity

None of the above related concepts exist in a topological space, where the notion of continuity is not quantifiable.

To define continuity in a topological space we operate not with a pair of nearby points $x_1,x_2$, but with a set of points; thus, the global picture emerges at once. "Preimage of an open set is open", a standard way to define continuity in a topological space, is naturally global. Equivalently, one can state it as "preimage of a closed set is closed". From here, your definition of continuity at a point, $$A \in \mathcal{P}(X), \, x_0\in \overline{A}\implies f(x_0)\in \overline{f(A)}\tag{1}$$ is obtained by localizing a global definition : you have localized the property of "containing all its limit points" by focusing on a particular limit point. As a result, (1) is more contrived than the global definition. If you try to reprove the basic results of point-set topology always using (1) as definition of continuity, you will likely find that the proofs become more cumbersome.

I prefer to read "not such a useful" as "not such a natural"; which however amounts to the same thing.

user127645
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