3
$\begingroup$

Since the definitions for a Hausdorff space and for a separated map is so similar, I wondered about the implications between them.

Suppose $X$ is a Hausdorff space, and we have a map $f:X \rightarrow Y$ to some topological space $Y$. So assuming that for two $x_1 \neq x_2 \in X$ we have that $f(x_1) = f(x_2)$, we know from the definition of a Hausdorff space that $f$ is separated.

Now onto the converse. Assuming that $f$ is separated. The Hausdorff condition on $X$ holds for all $x_1 \neq x_2 \in X$ when $f(x_1) = f(x_2)$, but since it's dependent on $f(x_1) = f(x_2)$ we cannot guarantee that the condition holds for all points $x_1 \neq x_2 \in X$. Is this correct?

Edit: Here's the definition of a separated map: A continuous map $f:X \rightarrow Y$ is called separated if any two points $x_1 \neq x_2 \in X$ with $f(x_1) = f(x_2)$ can be separated in the sense that there exists open sets $U_1, U_2 \subseteq X$ such that $x_1 \in U_1$, $x_2 \in U_2$ and $U_1 \cap U_2 = \emptyset$

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Hello, what is the definition of a separapled map? $\endgroup$ – Paul May 20 '16 at 12:34
  • $\begingroup$ @Paul I edited my post to include the definition. $\endgroup$ – Auclair May 20 '16 at 12:45
3
$\begingroup$

The notion of separated map is the relative version of the notion of a Hausdorff space. By relative one means that it's a statement about the fibers of a map $f\colon X\rightarrow Y$.

A continuous map $f\colon X\rightarrow Y$ is separated iff all fibers $f^{-1}(y)$ are Hausdorff topological spaces for the induced topology on $f^{-1}(y)$. In particular, $X$ is Hausdorff iff the continuous map from $X$ into the one-point topological space is separated.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see. Thanks for the answer. Are the arguments in my post correct though? $\endgroup$ – Auclair May 20 '16 at 12:55
  • 1
    $\begingroup$ Yes. You proved that any continuous map defined on a Hausdorff space is separated, and you indicated that the existence of a separated map $f\colon X\rightarrow Y$ does not imply that $X$ is Hausdorff. Indeed, take any nonHausdorff $X$. Then the identity map on $X$ is separated. $\endgroup$ – Johannes Huisman May 20 '16 at 13:07
  • $\begingroup$ Fibers Hausdorff under subspace topology does not imply separated. Consider $f : \{a,b,c\} \to \{0,1\}, a \mapsto 0, b \mapsto 0, c \mapsto 1$ where open sets of $\{a,b,c\}$ are subsets containing $c$ and $\{0,1\}$ has the indiscrete topology. Then $f^{-1}(0) = \{a,b\}$ is Hausdorff under subspace topology, as is $f^{-1}(1)$. But $(a,b)$ is in the closure of the diagonal in $\{a,b,c\}\times_{f}\{a,b,c\}$ and not in the diagonal. Hausdorff should be replaced with relatively Hausdorff to fix this. $\endgroup$ – user577413 Apr 17 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.