# Prove that $A_0$ and $B_0$ are separated subsets of $\mathbb{R}$; Exercise 21 Chapter 2, Baby Rudin

(Exercise 21 Chapter 2, Baby Rudin) I am trying to prove

Let $$A$$ and $$B$$ be separated subsets of some $$\mathbb{R}^k$$, suppose $$\textbf{a} \in A$$, $$\textbf{b} \in B$$ and define $$\textbf{p}(t) = (1-t)\textbf{a} + t\textbf{b}$$, for $$t \in \mathbb{R}$$. Put $$A_0 = \textbf{p}^{-1}(A), B_0 = \textbf{p}^{-1}(B)$$. [Thus, $$t \in A_o$$ iff $$\textbf{p}(t) \in A$$.]

Prove that $$A_0$$ and $$B_0$$ are separated subsets of $$\mathbb{R}$$. My attempt so far:

a. Assume to the contrary that $$\exists y$$ such that $$y \in A_0 \cap \overline{B_0}$$ which implies $$y \in A_0$$ and $$y \in \overline{B_0}$$. Then, $$\textbf{p}(y) \in A$$ and either $$y \in B_0$$ or $$y$$ is a limit point of $$B_0$$. If $$y \in B_0$$, then $$\textbf{p}(y) \in B$$ which would contradict that $$A$$ and $$B$$ are separated. If $$y$$ is a limit point of $$B_0$$, ...

My question: I am having trouble completing the proof. Can someone please suggest how this proof can be completed?

P.S. I found this proof but I have no idea why the idea of continuity was introduced in the first place, or even how one knows that $$p$$ is continuous, as the answer claims. I would like to complete this proof without using the concept of continuity, ideally, since Rudin hasn't introduced the concept of continuity so far (till Chapter 2).

Edit: We now claim that $$\mathbf{p}(t)$$ is continuous on all of $$\mathbb{R}$$.

Proof: Let $$\epsilon > 0$$ and $$c \in \mathbb{R}$$. Suppose $$\left|t-c\right| < \delta$$ where $$\delta = \frac{\epsilon}{\left|b-a\right|} > 0$$. Then, we have

$$\left|\mathbf{p}(t)-\mathbf{p}(c)\right| = \left|(1-t)\mathbf{a} + t\mathbf{b}-\mathbf{a}(1-c)-c\mathbf{b}\right| = (t-c)\left|\mathbf{b}-\mathbf{a}\right| < \frac{\epsilon}{\left|\mathbf{b}-\mathbf{a}\right|} \cdot \left|\mathbf{b}-\mathbf{a}\right| = \epsilon$$ and we are done.

Definition of a continuous function:

Suppose $$X, Y$$ are metric spaces, $$E \subset X, p \in E$$ and $$f$$ maps $$E$$ into $$Y$$. Then, $$f$$ is said to be continuous at $$p$$ if for every $$\epsilon > 0, \exists \delta > 0$$ such that $$d_Y(f(x), f(p))< \epsilon$$ for all points $$x \in E$$ for which $$d_X(x, p) < \delta$$

Definition of a closed set:

$$E$$ is closed if every limit point of $$E$$ is a point of $$E$$.

Definition of closure of a set (denoted by $$\bar{E}$$):

$$\bar{E} = E \cup E'$$ where $$E'$$ is the set of limit points of $$E$$.

Definition of a limit point

A point $$p$$ is a limit point of a set $$E$$ if every neighborhood of $$p$$ contains a point $$q \neq p$$ such that $$q \in E$$.

• It's a topological question, so continuity is helpful here. If you are not convinced that $p$ is continuous, you should prove it, it is not difficult.
– user17892
Commented Jun 21, 2020 at 15:10
• @JustinYoung Ok, I agree that $p$ is continuous. I added a proof for the same. But, how does the continuity of $p$ imply that $\overline{p^{-1}(A)}=p^{-1}(\overline A)$? Commented Jun 21, 2020 at 15:35
• Is it that I should have proven that $p^{-1}$ is continuous? Commented Jun 21, 2020 at 15:45
• $p^{-1}$ isn't a function so it can't be "continuous." Continuity is equivalent to $p^{-1}(U)$ is open for every open set $U$ is equivalent to $p^{-1}(F)$ is closed for every closed set $F$. Commented Jun 21, 2020 at 16:03
• $p$ is a function from $\mathbb R$ to $\mathbb R^k$ It is continuous on its domain $\mathbb R$. $p(c)$ does not make any sense. Commented Jun 21, 2020 at 16:15

Ok, here we go: this is a general proof of the following:

If $$p:X\to Y$$ is a continuous function and $$S\subseteq Y$$ is a subset, then $$\overline{p^{-1}(S)} \subseteq p^{-1}\left (\overline S \right )$$.

Proof: Assume that $$x \in \overline{p^{-1}(S)}$$. If $$x\in p^{-1}(S)$$, then clearly we are done since $$S\subset \overline S$$. If $$x\not\in p^{-1}(S)$$, then $$x$$ is a limit point of $$p^{-1}(S)$$. Consider $$p(x)$$. We want to show $$p(x) \in \overline S$$, then we need to show in this case that $$p(x)$$ is a limit point of $$S$$. Consider a neighborhood $$B(p(x), \epsilon)$$, for $$\epsilon >0$$. By continuity, there exists a $$\delta >0$$ such that if $$d(x,z) < \delta$$, then $$d(p(x), p(z)) < \epsilon$$ (I omit the decoration on the metrics for readability, do not assume the metrics are the same). Now, $$x$$ is a limit point of $$p^{-1}(S)$$, and $$B(x,\delta)$$ is a neighborhood of $$x$$, therefore, by definition there exists $$q\in B(x,\delta)\cap p^{-1}(S)$$. It follows that $$d(p(q),p(x)) < \epsilon$$, and $$p(q) \in S$$. We have now shown that $$p(x)$$ is a limit point of $$S$$. This completes the proof of $$\overline{p^{-1}(S)} \subseteq p^{-1}\left (\overline S \right )$$.

Applying this to your specific function, we conclude: $$\overline{A_0}\cap B_0 = \overline{p^{-1}(A)}\cap B_0 \subseteq p^{-1}\left (\overline A \right )\cap B_0 = p^{-1}\left (\overline A \cap B \right )= \emptyset$$, and by symmetry we get the corresponding inequality by switching the roles of $$A$$ and $$B$$.

Here is my counterexample for equality: Let $$k=2$$ and define $$A = \{(-1,0)\}\cup \{0\}\times (0,1]$$, and $$B = \{(1,0)\}$$, then let $$\textbf a = (-1,0)$$ and $$\textbf b = (1,0)$$. If we define $$p:\mathbb R \to \mathbb R^2$$ by $$p(t) = (1-t)\textbf a + t\textbf b$$, then you can verify that $$p^{-1}(A) = \{0\}$$, which is closed, so $$\overline{p^{-1}(A)}= \{0\}$$, but $$A' =\{(0,0)\}$$, so $$p^{-1}(\overline A) = \{0, 1/2\}$$. Thus, the inclusion $$\overline{p^{-1}(A)} \subseteq p^{-1}\left (\overline A \right )$$ is strict in this case.

• Thanks for the insightful answer! Commented Jun 22, 2020 at 5:28
• In the second-to-last line of the first paragraph, how do we know that $p(q) \in S$ and how does that tell us that $p(x)$ is a limit point of $S$? Commented Jun 22, 2020 at 17:23
• Well, $q\in p^{-1}(S)$ means by definition $p(q) \in S$. I showed that every neighborhood of $p(x)$ contains a point of $S$ (namely $p(q)$), and by assumption (recall that $x\not \in p^{-1}(S)$, so we know that $p(q) \not = p(x)$), this is the definition of $p(x)$ being a limit point of $S$.
– user17892
Commented Jun 22, 2020 at 18:21

This is a response to a query by the proposer in a comment to the A from @WilliamElliot.

Sets $$A,B$$ are separated iff $$A\cap \bar B=B\cap \bar A=\phi.$$ Sets $$A,B$$ are completely separated iff there exist disjoint open $$U,V$$ with $$A\subseteq U$$ and $$B\subseteq V.$$

If $$(X,d)$$ is a metric space and $$A, B$$ are separated subsets of $$X$$ then $$A, B$$ are completely separated.

PROOF: For each $$a\in A$$ take $$r_a\in \Bbb R^+$$ such that $$B\cap B_d(a,r_a)=\phi.$$ For each $$b\in B$$ take $$s_b\in \Bbb R^+$$ such that $$A\cap B_d(b,s_b)=\phi.$$

Let $$U=\cup_{a\in A}B_d(a,r_a/2)$$ and $$V=\cup_{b\in B}B_d(b,s_b/2).$$

To show $$U\cap V=\phi,$$ suppose instead that $$c\in U\cap V.$$ Take $$a\in A$$ such that $$c\in B_d(a,r_a/2).$$ Take $$b\in B$$ such that $$c\in B_d(b,s_b/2).$$ Then $$d(a,b)\le d(a,c)+d(c,b)

If $$K=r_a$$ then $$d(a,b) contrary to the def'n of $$r_a.$$

If $$K=s_b$$ then $$d(b,a) contrary to the def'n of $$s_b.$$

So $$c\in U\cap V$$ cannot exist.

• Thanks for clarifying! Commented Jun 22, 2020 at 5:28
• We can also prove it (eventually) by: (1). By a similar method, a metric space is normal ($T_4$)... (2). If $(X,d)$ is a metric space and $Y\subset X$ then the sub-space topology on $Y$ equals the $d$-metric topology on $Y,$ so $Y$ is $T_4$...(3). Theorem (elementary): If $X$ is a $T_4$ space then T.F.A.E.: (i). Every sub-space of $X$ is $T_4\,$ (ii). Every open sub-space of $X$ is $T_4\,$ (iii). Any 2 separated subsets of $X$ are completely separated. Commented Jun 22, 2020 at 16:28

Since $$A$$ and $$B$$ are seperated, there exist open disjoint $$U,V$$ with $$A \subset U, B \subset V$$. $$A_0 \subset K = p^{-1}(U), B_0 \subset L = p^{-1}(V)$$.
Show $$K$$ and $$L$$ are open and disjoint.

• I have posted, as an A, the explanation of why there exist disjoint open $U,V$ with $A\subset U$ and $B\subset V.$ Commented Jun 21, 2020 at 23:55