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Let $X$ be a topological space. It is known that if $\{x_n\}_{n \in \mathbb{N}}$ is a sequence in $X$ that converges to $x$, then the set of points $\{x,\{x_n\}_{n \in \mathbb{N}}\}$ is compact.

Is the similar statement for nets true? That is, if $X$ is a topological space and $\{x_{\lambda}\}$ is a net of points of $X$ that converges to $x$ (where $\lambda$ runs over some directed set), then the set of points $\{x,\{x_{\lambda}\}\}$ is compact?

If it is not compact, is it possible that it is relatively compact? (a set is called relatively compact if its closure is compact)

To make the question easier, we can assume that $X$ is Hausdorff.

Any help would be greatly appreciated.

Pawel
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3 Answers3

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Suppose the directed set you use as indices is $\mathbb{Z}$, and the net $\mathbb{Z} \to \mathbb{R}$ sends $n \mapsto 2^{-n}$. Then the net converges to 0; however, the image of the net along with its limit is $\{ 2^n \mid n \in \mathbb{Z} \} \cup \{ 0 \}$ which is unbounded so it cannot be compact or even relatively compact.

Daniel Schepler
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Let the directed set be (0,1) and the net assign x to x in R.
This net converges to 1 and its image along with 1 is (0,1]
which is not compact.

William Elliot
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No. Let the directed set $\Lambda$ consist of the natural numbers, ordered as usual, and infinitely many additional elements, incomparable with each other but each $<$ all of the natural numbers. Then a net indexed by $\Lambda$ converges iff the sequence of its values at the natural numbers converges, while the values at the additional elements are completely arbitrary. Those arbitrary values can be chosen to make the set of all the net's values plus the limit non-compact.

Andreas Blass
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