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If $X$ be a metrizable space, then it is known that "limits of sequences suffice to determine the topology."

Maybe this is a basic question, but how can we identify the open sets of the topology from those limits?

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  • $\begingroup$ $U$ is open in The topology iff the complementary of $U$ in $X$ contains the limit of each converging sequence of elements of the complementary of $U$. $\endgroup$ – Clément Guérin Jan 19 '16 at 10:54
  • $\begingroup$ Three almost identical answers over the span of three minutes! Yay! $\endgroup$ – Najib Idrissi Jan 19 '16 at 11:03
  • $\begingroup$ I had to think on that, wops :) thanks everybody $\endgroup$ – Paolo Leonetti Jan 19 '16 at 11:29
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A set $A$ is closed if and only if it contains the limit of each convergent sequence of points of $A$, so in principle we can identify the closed sets. The open sets are just the complements of the closed sets, so we get them as well.

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For each $A\subseteq X$, $A$ is open if and only if for every $x\in A$ and every $\{x_n\}\to x$, there exists some $N$ such that $x_n\in A$ for all $n\geq N$.

Essentially this says "a set is open if every sequence that converges in the set has its tail in the set."

Note: While this does determine the topology, it tends to be unwieldy.

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A set is closed if and only if all convergent sequences with elements from the set have a limit which is also in the set.

Therefore, knowing all limits, you can also determine the closure of every set, therefore you can also determine all closed sets (and with them, the topology).

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I haven't really studied any of this formally, but I imagine the following are true:

Let $X$ denote a topological space and $A$ denote a subset thereof. Then:

  • $A$ is closed iff for all nets $x : A \leftarrow D$ and all $x_0 \in X$, we have that if $x$ converges to $x_0$, then $x_0 \in A$.

  • $A$ is open iff for all $a \in A$ and all nets $x : X \leftarrow D$, it holds that if $x$ converges to $a$, then $x$ is eventually in $A$ (i.e. there exists $i \in D$ such that for all $j \geq i$, we have $x_j \in A$.)

If $X$ is furthermore assumed to be a sequential space, you should be able to replace the word "net" by "sequence" in all of the above characterizations, without issue.

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