If $\phi$ is a net that converges to $x$ then $x$ is the only point of accumulation. I'm trying for contradiction but I can't see it.

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    $\begingroup$ For this statement you need that the space is Hausdorff. For instance, if the space has the indiscrete topology, every net converges to every point. $\endgroup$ – Ulli Sep 16 at 6:35

We need that $X$ is $T_2$ (Hausdorff). Suppose that $p \neq x$ is an accumulation point of the net $(x_i)_{i \in I}$ that converges to $x$.

Let $U$ and $V$ be disjoint open sets such that $x \in U$ and $p \in V$ by Hausdorffness.

Then as $x_i \to x$ we have an index $i_0$ such that

$$\forall i \ge i_0: x_i \in U$$

As as $p$ is an accumulation point of $(x_i)_i$ we have a $j \ge i_0$ such that $x_j \in V$. But then this $x_j \in U \cap V$ , contradicting the disjointness.

If $X$ is not Hausdorff there are two points $x \neq y$ and a net $(x_i)_i$ in $X$ that converges to both $x$ and $y$, so then $y$ is another accumulation point besides the limit. So Hausdorff is necessary and sufficient for this fact to hold.

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