Trying to follow proof in Royden that every metric space is normal

Question

In Section $11.2$: The Separation Properties in Royden's Real Analysis, there is a proposition that states that "every metric space is normal".

As a reminder of what a normal space is, it is a topological space that satisfies the normal separation property, which is the following:

The Normal Separation Property: The Tychonoff separation property holds and moreover, each two disjoint closed sets can be separated by disjoint neighborhoods (According to Royden, Tychonoff is when for every two points $x$ and $y$ in $X$, there exists a neighborhood of $x$ that does not contain $y$ and a neighborhood of $y$ that does not contain $x$).

The proof goes as follows:

Let $(X,\rho)$ be a metric space. Define the distance between a subset $F$ of $X$ and point $x$ in $X$ by $dist(x,F)= \inf\{ \rho(x, x^{\prime})|x^{\prime} \in F \}$.

Let $F_{1}$ and $F_{2}$ be closed disjoint subsets of $X$. Define

$\mathcal{O}_{1} = \{x \in X\,|\,dist(x,F_{1})<dist(x,F_{2})\}$ and $\mathcal{O}_{2} = \{x \in X\,|\,dist(x,F_{2})<dist(x,F_{1})\}$.

Up to this point, I'm fine. But then, it says

Since the complement of a closed set is open, $dist(x,F)>0$ if $F$ is closed and $x$ does not belong to $F$.

I think the reason why this is true is because if $x \notin F$, and $F$ is closed, then $x$ cannot be a point of closure of $F$. So, then, there must exist at least one neighborhood of $x$ that does not contain a point of $F$, so the infimum of the metric distance between $x$ and any point of $F$ must be strictly positive. But, I'm not entirely sure that's correct reasoning.

Also, it says that $F_{1} \subseteq \mathcal{O}_{1}$, $F_{2} \subseteq \mathcal{O}_{2}$, which I have a vague intuitive idea must be true, but I couldn't put it into words if you forced me to. So, if I could get a better explanation of why this is, that would be great. By the way $F_{1}$ and $F_{2}$ are defined, though, it's obvious that $\mathcal{O}_{1} \cap \mathcal{O}_{2} = \emptyset$.

The last part of the proof is the part that is really bugging me and that I really need help understanding:

Moreover, using the triangle inequality for the metric, it is not difficult to see that $\mathcal{O}_{1}$ and $\mathcal{O}_{2}$ are open.

To do that, I know we would have to show that we could construct a ball around an arbitrary point $x_{1} \in \mathcal{O}_{1}$ that is contained entirely within $\mathcal{O}_{1}$ and likewise for some $x_{2} \in \mathcal{O}_{2}$. If $y_{1}$ is some other point in $B_{r}(x_{1})$, then, I'd need to show that the metric distance (meaning $\rho$ of something) between $y_{1}$ and an arbitrary point $x \in \mathcal{O}_{1}$ is less than or equal to $\rho(y_{1},x_{1}) +\,$ metric distance between $x_{1}$ and $x$.

The problem is is that $\mathcal{O}_{1}$ isn't necessarily a nice round ball with a nice constant radius $r$, so even once I've got the triangle inequality set up, how do I represent the metric distance between $y_{1}$ and $x$, or between $x_{1}$ and $x$? I guess I'm just so used to seeing these kinds of problems written in terms of balls that I have no idea how ones that don't necessarily have to do with balls should be written.

Could somebody please let me know what this triangle inequality should look like?

Answer 1

I think the reason why this is true is because if $x∉F$ and $F$ is closed, then $x$ cannot be a point of closure of $F$. So, then, there must exist at least one neighborhood of $x$ that does not contain a point of $F$, so the infimum of the metric distance between $x$ and any point of $F$ must be strictly positive. But, I'm not entirely sure that's correct reasoning.

This reasoning is actually correct. We'll have ${\rm dist}(x,F) \geq r>0$, where $r$ is the radius of the ball that keeps $x$ away from $F$.

Also, it says that $F_{1} \subseteq \mathcal{O}_{1}$, $F_{2} \subseteq \mathcal{O}_{2}$, which I have a vague intuitive idea must be true, but I couldn't put it into words if you forced me to. So, if I could get a better explanation of why this is, that would be great. By the way $F_{1}$ and $F_{2}$ are defined, though, it's obvious that $\mathcal{O}_{1} \cap \mathcal{O}_{2} = \emptyset$.

If $x \in F_1$, then $x \in {\cal O}$ because $0 = {\rm dist}(x,F_1) \color{red}{<} {\rm dist}(x,F_2)$, and the inequality is strict because $F_1$ and $F_2$ are disjoint.

Moreover, using the triangle inequality for the metric, it is not difficult to see that ${\cal O}_1$ and ${\cal O}_2$ are open.

An easy way to see this is to prove that if $F$ is any set $x \mapsto {\rm dist}(x,F)$ is continuous, and if $x \in {\cal O}_1$, then ${\rm dist}(x,F_2)-{\rm dist}(x,F_1) > 0$, and we get the ball using continuity and conservation of sign. The proof of continuity of $x \mapsto {\rm dist}(x,F)$ uses the triangle inequality in the form $|{\rm dist}(x,F) - {\rm dist}(y,F)|\leq d(x,y)$, and then choosing $\delta = \epsilon$.

An easier proof of the fact that every metric space is normal uses the continuity of the function above to define: $$f(x) = \frac{{\rm dist}(x,F_1)}{{\rm dist}(x,F_1)+{\rm dist}(x,F_2)},$$and taking the open sets to be the inverse images of $[0,1/2[$ and $]1/2,1]$. The function is well-defined because the closed sets are disjoint.

Answer 2

To see that $O_1$ is open: For $p\in O_1,$ for brevity let $A=d(p,F_1)$ and $B=d(p,F_2).$ Let $ r=(B-A)/3.$ Then $r>0.$

We have $B(p,r)\subset O_1.$

Proof: For any $s\in B(p,r)$ and any $q_2\in F_2$ we have $$d(q_2,s)\geq d(q_2,p)-d(s,p)\geq B-d(s,p)>B-r= (2 B+A)/3.$$ $$\text {Hence }\; d(s,F_2)\geq (2 B+A)/3.$$ But for some $q_1\in F_1$ we have $d(p,q_1)< A+r$, so we have $$d(s,q_1)\leq d(s,p)+d(p,q_1)<d(s,p)+(A+r)<r+(A+r)=(2 B+A)/3.$$ Hence $d(s,F_1)<(2 B+A)/3\leq d(s,F_2).$ So $s\in O_1.$

Remark :$(B-A)/3$ is the largest value of $r$ we can use for this method. Any $r'\in (0,(B-A)/3]$ will work. The idea is that if $p$ is closer to $F_1$ than to $F_2$ then any $s$ sufficiently near to $p$ will also be.