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A short and hopefully simple question for someone with more experience in topology:

If a topology is induced by a mode of convergence and in fact nothing more is known apriori, whether this topology is metrizable, first-countable or anything else? Is continuity then equivalent to sequential continuity?

I'm quite sure this is should be true, but topology is tricky and my intuition not too developed in this field.

Mr. Barrrington
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2 Answers2

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Let be X and Y topological spaces and $f : X \rightarrow Y$ is a map.

Then it holds:

1) If $f$ is continuous then it is sequentially continuous.

2) If $f$ is sequentially continuous and $X$ is first countable, then $f$ is continuous.

The prove of 1) is easy. You just have to use the definition.

For the prove of 2) you have to use that: $f$ is continuous => $f(\overline{M}) \subseteq \overline{f(M)}$

So there is equivalence if and only if $X$ is first countable.

Siminore
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Marm
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  • Ok, thx for this input balla! But are topologies given by means of convergence always first countable? Or not? I mean if I more or less define what it means for a set to open by sequences, why shouldn't continuity and sequential continuity coincide in this case? – Mr. Barrrington Jul 01 '14 at 13:43
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    You only proved that there is equivalence if $X$ is first countable. The other implication doesn't hold. It is enough to assume that $X$ is sequential. – user87690 Jul 01 '14 at 17:15
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There are sequential spaces, these are topological spaces such that a set $A$ is closed if $A$ is sequentially closed, meaning $A$ contains the limits of all sequences in $A$. One can say that a sequential space has the final topology with respect to all continuous maps from $\hat{\Bbb N}$, the one-point-compactification of $\Bbb N$, to $X$.
It's not difficult to show that $X$ is sequential if and only if every sequentially continuous function $f:X\to Y$ is continuous.
A less general class of spaces is the so-called Frechet-Urysohn spaces (FU). A space is FU if a limit point $x$ of $A$ is always the limit of some sequence within $A$. These spaces include the first-countable spaces.

If $X$ is FU and $f:X\to Y$ is pseudo-open, then $Y$ is FU, too. Since every closed surjection is pseudo-open, the FU property carries over to quotients of $X$ by a closed subspace $A$. For example, $\Bbb R/\Bbb Z$ is FU. Note that this space is not first-countable (at $\Bbb Z$).

Stefan Hamcke
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  • ok, so this was what I somehow suspected. By getting a topology via a mode of convergence I de facto define the space to be sequential. And thus sequential cont. is equivalent to continuity – Mr. Barrrington Jul 03 '14 at 09:03