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I was wondering whether a net always has a countable subnet? Since there is a criterion for continuity using nets, and in some spaces we can check only for sequences. It would seem to me that a convergent net to a point with a countable local base we can find, but is this even true?

Keen-ameteur
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1 Answers1

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No. The net on $\omega_1$ into $\omega_2,$ sending each
countable ordinal to itself is a counter example.

If a net converges to a point p with countable
local base, then yes, there is a countable subnet.

Let { $U_1, U_2,$ ... } be the countable base and use
the decending local base { $U_1, U_1 \cap U_2,$ ... }
to construct a countable subnet.

William Elliot
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  • You mean with the order topology, right? – Keen-ameteur Feb 05 '18 at 19:38
  • Yes, with the usual topology for the ordinals. @Keen-ameteur – William Elliot Feb 05 '18 at 19:52
  • The general definition of a subnet is the first in the: [[https://math.stackexchange.com/questions/1126609/different-definitions-of-subnet]] – Darman Nov 23 '19 at 12:13
  • If $(0,1]$ is taken with the reverse order, then $(0,1]\times \omega_1$ is taken with the product order is a directed set. If $\phi:\omega_1\to[0,2\pi)$ is a bijection, then $x:(0,1]\times\omega_1\to\mathbb{C}$ defined by $x(a,b)=ae^{i\phi(b)}$ is a net that converges to zero in the usual topology of $\mathbb{C}$. This net doesn't have a countable subnet even though it goes to a metric space (in particular with countable local base) and converges. – conditionalMethod Nov 24 '19 at 12:50
  • @William Elliot; When X is with countable local base, 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 13:54