A well-known theorem is: If $f:X \to Y$ is a quotient map whose domain $X$ is a compact $T_2$-space and if the equivalence kernel $\operatorname{ker} f = \bigl\{(x, u) \in X \times X : f(x)=f(u)\bigr\}$ is closed in $X \times X$, then the quotient space $Y$ is also $T_2$.
Stated in an equivalent way, the theorem is: If $\sim$ is an equivalence relation on a compact $T_2$-space $X$ and if $\sim$, that is, $\bigl\{(x, u) \in X \times X : x \sim u\bigr\}$, is closed in $X$, then the quotient space $X/\sim$ is also $T_2$.
Consider this example: $X = [-1, 1]$ with its usual topology; $Y = X/\sim$ where $\sim$ is the equivalence relation defined by $x \sim u$ if and only if $u = t x$ for some $t \neq 0$; and $f$ is the quotient map.
The space $Y$ consists of just two points, namely, the equivalence class $[0] = \{0\}$ of $0$ and the equivalence class $[1] = [-1, 1] \setminus \{0\}$.
Question: What is wrong with the following two apparently inconsistent assertions?
- The equivalence kernel $\operatorname{ker} f = [-1, 1] \times [-1, 1]$, which is closed in $X \times X$, so by the theorem, $Y$ is also $T_2$.
- The only nonempty proper open subset of $Y$ is the singleton $[1]$, so $Y$ cannot be $T_2$ (nor even $T_1$).
Evidently something is wrong, but I'm just not seeing it!