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Let $(X,\tau)$ be a topological space, let $x^*$ be an element of $X$, and let $(x_{\alpha})$ be a net from some directed set $A$ into $X$, that converges toward $x^*$. Is there necessarily some sequence $(x^1, x^2, \dots)$ in the net (i.e. in the set $\{x_{\alpha}\}_{\alpha\in A}$) that converges to $x^*$?

Here's my attempt at solving this question. Am I in the right direction?

Suppose $\tau$ is metrizable. Choose some metric $d:X\times X\rightarrow[0,\infty)$ that induces $\tau$. For every $n \in \{1,2,\dots\}$ denote by $S_n$ the sphere about $x^*$ with radius $\frac{1}{n}$. For every $n \in \{1,2,\dots\}$ $S_n$ is a neighborhood of $x^*$ in $\tau$ and therefore (since we are given that $(x_{\alpha})$ converges to $x^*$) there exists some $\alpha_n \in A$ such that $x_{\alpha_n} \in S_n$; define $x^n := x_{\alpha_n}$. The sequence $(x^1, x^2, \dots)$ converges to $x^*$ in the metric space $(X,d)$, and therefore in the original topological space. But what if $\tau$ isn't metrizable?

Evan Aad
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  • You mean a subnet that is also a sequence (which is a net defined on $\mathbb{N}$) to be more precise, instead of "a sequence in the net". – Henno Brandsma May 24 '18 at 23:13

2 Answers2

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No. For example within [0,$\omega_1$] there is
a net converging to $\omega_1$ but no sequence.
Yes if the space is first countable.

William Elliot
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  • Thanks. Can you please elaborate on your first point? – Evan Aad May 24 '18 at 20:55
  • You were the first to answer, and you gave a correct answer. However, since you didn't provide further details, as per my request, I was not able to verify your answer. It is for this reason that I've marked the other answer as the accepted answer. But I've given you an upvote. – Evan Aad May 25 '18 at 10:18
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Your argument for metric spaces can be generalised to first countable spaces and then shows the following (well-known):

if $f: A \to X$ is a net into a first countable space $X$, such that $f \to x$, then there is a subnet that is a sequence (subsequence) $f': \mathbb{N} \to X$ that also converges to $x$.

For general spaces such a thing cannot be said: $X = \omega_1 +1$ (the successor of the first uncountable ordinal ) has a net $\omega_1 \to X$ (send $x$ to $x$) that converges to the point $\omega_1$, but no subnet that is a sequence can converge to that point, as $\omega_1$ has uncountable cofinality (every countable subset is bounded above).

Henno Brandsma
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  • In the terminology I've seen, a "subnet" of $A \to X$ is induced by a dominant map $B \to A$ - with the point being to set up the definition so that if a net converges to $L$ then every subnet also converges to $L$. So, with this terminology, it would in fact be impossible to have a sequence which is a subnet of your net $\omega_1 \to X$. – Daniel Schepler May 24 '18 at 23:21
  • @DanielSchepler Yes, that's the point of the example, that there aren't even subnets that are sequences, as we can find in the case of a convergent net in a first countable space. – Henno Brandsma May 24 '18 at 23:23
  • Thanks. However, I'm a little rusty on my set theory. In your counterexample, what is the topology on $X$? – Evan Aad May 25 '18 at 07:00
  • @EvanAad the order topology, as is usual on ordinal spaces. – Henno Brandsma May 25 '18 at 07:28
  • Would you mind fleshing out the details a little? What are the properties of the underlying set and of the topology on it that you implicitly use in your counterexample? You don't need to prove them, just state them, please. – Evan Aad May 25 '18 at 07:57
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    @EvanAad start by reading this and the Wikipedia page and that should cover it. – Henno Brandsma May 25 '18 at 08:04
  • Thanks. The document at the first link was illuminating. – Evan Aad May 25 '18 at 10:18
  • @Henno Brandsma; Your answer is false. because of: https://math.stackexchange.com/questions/3448778/does-every-net-in-mathbbc-converging-to-0-have-a-countable-subnet – Darman Nov 24 '19 at 13:45