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While working on a takehome for my functional analysis course I stumbled upon this small lemma

A net $(x_i)_{i\in I}$ in a topological space $X$ converges to a point $x\in X$ if and only if every subnet has a accumulation point in $x$.

This is a slightly stronger formulation of the following well known result in topology.

A net $(x_i)_{i\in I}$ in a topological space $X$ converges to a point $x\in X$ if and only if every subnet converges to $x$.

I managed to come up with the following proof, but I doubt my judgement because it seems a little unbelievable for me to come up with a stronger version of an existing mathematical result. Can you check my proof?

the implication from left to right is trivial because if $(x_i)_{i\in I}$ converges to $x$ then so will any subnet. Convergence to $x$ implies that the subnet has a accumulation point in $x$ as well, because this is a weaker statement.

Now if $(x_i)_{i \in I}$ does not converge to $x$, it has a subnet which does not converge to $x$, $(x_{\sigma(j)})_{j\in J}$. This means there is an open neighbourhood $U$ of $x$ such that for any $j \in J$ there exists a $j' \geq j$ such that $x_{\sigma(j')} \not\in U$. Using the map $$J \to J : j \to j'$$ we find the subnet $(x_{\sigma(j')})_{j \in J}$ which has no accumulation point in $x$.

Saajmen
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  • You need subnet of subnet, because if every subnet converges to $x$ then it's trivial that the original net converges to $x$ too. That's almost void. – Henno Brandsma Dec 21 '21 at 08:55
  • The second statement from right to left is void now. See previous comment. – Henno Brandsma Dec 21 '21 at 09:05
  • The final proof attempt is false of course. A logical fallacy. – Henno Brandsma Dec 21 '21 at 15:40
  • The second statement is true, I don't know what you mean with "void". Can you elaborate? – Saajmen Dec 21 '21 at 15:46
  • If we assume the right hand side the left is immediate as the net is a subnet of itself. – Henno Brandsma Dec 21 '21 at 15:50
  • correct, it's trivial but still true. I don't know why this would pose a problem. – Saajmen Dec 21 '21 at 15:51
  • It adds nothing to a possible proof of the original lemma, while you claim it does. That’s a problem IMO – Henno Brandsma Dec 21 '21 at 16:04
  • If the net does not converge to $X$ then it has a subnet that does not converge to $x$. Agreed: the net itself. But this does not give any proper subnet. And $x$ could still be an accumulation point for the net. So the proof halts; it’s nonsense. – Henno Brandsma Dec 21 '21 at 16:13
  • I construct a proper subnet with the following procedure:

    This means there is an open neighbourhood U of x such that for any j0∈J there exists a j≥j0 such that xσ(j)∉U. In other words, we found a subnet without an accumulation point in x.

    Because for any $j \in J$ we can find this $j' \in J$ such that $j'\not\in U$. the net $(x_{\sigma(j')})_J$ now does not have an accumulation point in $x$.

     – Saajmen Dec 21 '21 at 16:16
  • You construct nothing at all. Where’s the definition? – Henno Brandsma Dec 21 '21 at 16:17
  • i'll elaborate on the construction in the original body of the question – Saajmen Dec 21 '21 at 16:20
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    $\sigma$ is superfluous or define the subnet by $\sigma(j)=j’$ using AC. You still need an argument as to why it’s a subnet (depending on the definition of subnet; there are several). – Henno Brandsma Dec 21 '21 at 16:25
  • Yes I see that $\sigma$ is not needed, thanks for your input. I didn't know there were conflicting definitions for a subnet, I guess this works for the definition we are using at the moment. – Saajmen Dec 21 '21 at 16:28
  • Which is? You don’t specify. – Henno Brandsma Dec 21 '21 at 16:30

2 Answers2

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The statements are trivially equivalent if you know the standard fact:

The net $(x_j)_j$ has $x$ as an accumulation point iff it has a subnet that converges to $x$.

So the second version of the statement (which you claim to know) is immediately equivalent to your lemma.

Henno Brandsma
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The question is a duplicate so I will close it. Thank you for your help Henno.

Every subnet of $(x_d)_{d\in D}$ has a subnet which converges to $a$. Does $(x_d)_{d\in D}$ converge to $a$?

Saajmen
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