Here is Prob. 1, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint.

Here is the definition of regular space:

A topological space $X$ is said to be regular if

(i) $X$ satisfies the $T_1$ axiom (i.e. singleton subsets and hence finite subsets of $X$ are closed in $X$), and

(ii) for every point $x \in X$ and every closed set $B$ in $X$ not containing $x$, there are disjoint open sets $U$ and $V$ in $X$ such that $x \in U$ and $B \subset V$.

Moreover, according to the usage of Munkres, in any topological space $X$ and for any point $x\in X$, any open set $U$ of $X$ is said to be a neighborhood of $x$ if $x \in U$.

My Attempt:

Suppose that topological space $X$ is regular. Let $x$ and $y$ be any two distinct points of $X$. Let us put $B \colon= \{ y \}$.

Now as $x \in X$ and as $B$ is a closed set in $X$ not containing $X$, so by the regularity of $X$ there exist disjoint open sets $U$ and $V$ in $X$ such that $x \in U$ and $B \subset V$, that is, $x \in U$ and $y \in V$. Note that here we have defined set $B$ to be the singleton set $\{ y \}$.

Now as $X$ is a regular space, as $x$ is a point of $X$, and as $U$ is a neighborhood of (i.e. an open set containing) $x$ in $X$, so by Lemma 31.1 (a) in Munkres there exists a neighborhood $U^\prime$ of $x$ such that $\overline{U^\prime} \subset U$.

Similarly, there exists a neighborhood $V^\prime$ of $y$ in $X$ such that $\overline{V^\prime} \subset V$.

Finally, as $\overline{U^\prime} \subset U$ and $\overline{V^\prime} \subset V$, and as $U$ and $V$ are disjoint, so $\overline{U^\prime}$ and $\overline{V^\prime}$ are also disjoint.

Thus for any two distinct points $x, y \in X$ we have neighborhoods $U^\prime$ and $V^\prime$ of $x$ and $y$, respectively, such that the closures $\overline{U^\prime}$ and $\overline{V^\prime}$ are disjoint.

This proof and this one are very similar.

Is this proof correct and clear enough? Or, are there lacks and gaps in it?

Finally, what about the converse?


This is the same proof as the normality one, in essence. Separate points by open sets first $x \in U_x$, $y \in U_y$ say. This can be done as $X$ is regular (and $T_1$ is used in that $\{x\}$ and {y}$ are closed sets).

Inside $U_x$ and $U_y$ we can apply 31.1(b) to each and get $V_x$ open containing $x$, $V_y$ open containing $y$ with $\overline{V_x} \subseteq U_x$ and $\overline{V_y} \subseteq U_y$ and then $V_x$ and $V_y$ are as required as already $U_x \cap U_y = \emptyset$. All this is essentially your proof as well, so that is also OK. My write-up is just a bit more concise, but verbosity is a common theme in your posts.

The $\Bbb R_K$ topology is an example of a space where we can separate any two points by open sets with disjoint clsoures (sometimes called an "Urysohn" or strongly Hausdorff space) but which is not regular. Not part of exercise 1 but always good to ask oneself.


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.