Let $X=\{0,1\}$ and $\tau =\{ \emptyset , X, \{0\} \}$. Is $(x_n)=0,0,... $ convergence?

Question

Let $X=\{0,1\}$ and $\tau =\{ \emptyset , X, \{0\} \}$. And $(x_n)=0,0,... $.

Is $(x_n)$ convergence ?

$(X,\tau)$ is not a Hausdorff.

$(x_n)$ converges to $0$ since:

the neighborhoods of $0$ are $\{0\}$ and $X$, all of them contain all terms of the sequence.

$(x_n)$ converges to $1$ since:

the only neighborhood of $1$ is $X$, which is contain all terms of the sequence.

So $(x_n)$ convergence for all $x\in X$. Is that true, please?

Answer 1

You did it right. What you probably meant at the end is that for all $x\in X$, $x$ is a limit of the sequence $(x_n)$, which is what you just proved.

That's an example that uniqueness of the limit fails for general non-Hausdorff spaces.