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 $x$.

## Best Answer

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

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