# Points of acumulations in a net [closed]

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.

• 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. – 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.