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I saw a statement on http://en.wikipedia.org/wiki/Metric_space#Continuous_maps:

f is continuous if and only if it is continuous on every compact subset of M1.

It is stated for the case that f is a mapping between two metric spaces.

I was wondering if the statement is true for mappings between general topological spaces? Why?

Thanks and regards!


More questions:

What makes the statement true for a mapping between two metric spaces? Can you give a sketch for the proof?

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  • $\begingroup$ This is probably in the link that Qiaochu provided, but I thought it may as well be mentioned here. Continuity on compact subsets implies sequential continuity, because a convergent sequence and its limit point form a compact set. Sequential continuity is equivalent to continuity for first countable spaces. $\endgroup$ – Jonas Meyer Oct 29 '10 at 1:24
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The statement is false in general. Let $X$ be an uncountable set with the cocountable topology. Then all the countable subsets of $X$ have the discrete topology and are therefore closed but not compact, so the only compact subsets of $X$ are the finite subsets, which also carry the discrete topology. It follows that every function $X \to Y$, where $Y$ is any topological space, is continuous on compact subsets of $X$. (But, for example, if $Y$ has the same underlying set with the discrete topology, the "identity" function $X \to Y$ is not continuous.)

Edit: In response to your second question, the property you are looking at is equivalent to being compactly generated. A topological space $X$ is compactly generated if it satisfies the following condition: a subset of $X$ is open if and only if its intersection with every compact subset is open.

So why are metric spaces compactly generated? Recall that a space is first-countable if every point has a countable neighborhood basis. Metric spaces are first-countable because at any point the sequence of open balls of radius $\frac{1}{n}, n \in \mathbb{N}$ forms a countable neighborhood basis. And first-countable spaces are always compactly generated; see the proof here.

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    $\begingroup$ Thanks! What makes the statement true for a mapping between two metric spaces? Can you give a sketch for the proof? $\endgroup$ – Tim Aug 18 '10 at 6:51
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Let $X$ be a topological space. Then $X$ has the property that a function $f:X\to Y$ (where $Y$ is an abritrary topological space) is continuous if its restriction to each compact subspace is continuous if and only if $X$ is compactly generated topological space.

The space $X$ is compactly generated if a subset $A$ whose intersection with any compact subspace $K$ of $X$ is open in $K$, is open in $X$.

Most familiar topological spaces are compactly generated (Euclidean spaces, manifolds, CW-complexes etc.) but not all; see Qiaochu's answer for an example.

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show that a subset of a metric space X is closed iff its intersection with every compact subset of X is closed

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