Let $X = [0,1]$ and $\sim$ have the following equivalence relation over $X$: is the quotient first countable?

Question

Let $X = [0,1]$ and $\sim$ have the following equivalence relation over $X$:

$$x\sim y\iff x = y\text{ or }x, y \in {]}0, 1{[}$$

Show that $X/{\sim}$ is not first countable.

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First countable means that for any distinct $x, y\in X$, there are $U, V \subset X$ open such that $x\in U, y\in V, x\notin V$ and $y\notin U$.

And

$[x] = \{y\in X : y\sim x\}$

$X/{\sim} = \{[x]: x\in X\}$

I know it should be a simple exercise, but I stuck, any help would be most welcome. Thank you in advance for any assistance.

Answer 1

$X/\sim$ has exactly 3 points and thus it is first countable. Just like any space with finitely many points or more generally countable number of open subsets.

First countable means that for any distinct $x, y\in X$, there are $U, V \subset X$ open such that $x\in U, y\in V, x\notin V$ and $y\notin U$.

No, this is not the definition of "first countable". What you've described is being $T_1$.

And indeed, $X/\sim$ is not $T_1$. First of all note that $X/\sim=\{[0],[1/2], [1]\}$ has $3$ points. Furthermore proper, nontrivial open subsets are precisely $\{[0],[1/2]\}$, $\{[1/2]\}$ and $\{[1/2],[1]\}$. So can you separate say $[0]$ from $[1/2]$?

More generally a finite space is $T_1$ if and only if it is discrete. And our $X/\sim$ is not discrete: neither $[0]$ nor $[1]$ are open.