I am trying to prove the above. More specifically I am trying to prove the direction "if all subnets of a given net have limits then the net in question has a limit"

The definition I am using for a subnet is as follows (From Folland)

A subnet of a net $(x_\alpha)_{\alpha\in A}$ is a net $(y_{\beta})_{\beta\in B}$ together with a map $h:B\rightarrow A$ such that

-For every $\alpha_{0}\in A$, there exists $\beta_{0}\in B$ such that $\forall \beta$ such that $\beta_{0}\preccurlyeq \beta$ we have $\alpha_{0}\preccurlyeq h(\beta)$


I feel the only way I can approach this is via contradiction. Suppose that $(x_\alpha)_{\alpha\in A}$ did not have a limit, then...then what. I feel I need a way to talk about convergence without knowing what the limit is. In real analysis, one could do this using Cauchy sequences. A quick look on wikipedia shows that a thing called a Cauchy net exists, but since my lecturer gave us this problem without ever talking about Cauchy nets I feel there should be a way to solve this without referring to Cauchy nets.

To be clear, the only things we were taught was basic point set topology and the definition of a net and a subnet, so I would like, if possible, an answer that only utilises these things. A hint or complete answer is welcome.

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    $\begingroup$ What about the fact that any net $(x_\alpha)_{\alpha \in A}$ is a subnet of itself, via the identity map $A \to A$? $\endgroup$ – Daniel Schepler Dec 11 '18 at 1:45
  • $\begingroup$ My life would be so much easier if I wasn't an idiot, but thank you for taking the time to point this out to me. $\endgroup$ – Damo Dec 11 '18 at 1:49
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    $\begingroup$ I guess correctly specified version of your question is "if every subnet of a given net has a convergent subsubnet with common limit $L$, then the net in question converges to $L$." Certainly, this statement has a sequence counterpart, and it is obviously true for any sequence. It turns out that it is also true for any net: see math.stackexchange.com/questions/897957/on-nets-convergence. $\endgroup$ – Song Dec 11 '18 at 2:00

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