Is this the correct conclusions about a quotient topology on $\mathbb R$?

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Can someone tell me if my conclusions about this space is correct?

I have an equivalence relation for $x,y \in \mathbb R$ given by $$x\sim y \Longleftrightarrow (x = y) \lor (x,y \in (-a,a])$$ for some fixed $a > 0$.

The equivalence class singleton given by $\{[0]\}$ is not closed. It is not closed because $(-a,a]$ is not closed in $\mathbb R$. This implies that the quotient topology is not $T_2$.

Every other equivalence class singleton is closed because every other equivalence class corresponds to one point and singletons of points are closed in $\mathbb R$.

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Yes, that's correct. The singleton $\{[0]\}$ is not closed (it is not open either) since $(-a,a]$ is not closed (and not open). This actually shows that your quotient space is not $T_1$. In particular, it's not $T_2$, so you were correct.

The quotient space is $T_0$, though, for if $v,w$ are equivalence classes, then at least one of them is closed.

If I'm not mistaken, then your quotient space is also normal. The proof is not so difficult, but requires a few lines of reasoning. If you are interested, I can provide some hints, or a complete proof, if you want.