# Sigma compactness question, diagonal open implies $X$ is discrete and boundary exercise

some questions:

Question 1. It is a standard fact that $\mathbb{R}^{\omega}$,endowed with the product topology, is not the continuous image of $\mathbb{R}$.

Basically the proof consists in showing that $\mathbb{R}^{\omega}$ is not $\sigma$-compact.

What are other ways of arguing that $\mathbb{R}^{\omega}$ is not the cts image of $\mathbb{R}$?

I was thinking in using local compactness but I don't think this works, certainly $\mathbb{R}^{\omega}$ is not locally compact but in general local compactness is not preserved under continuous maps.

Question 2. Show that the diagonal in $X \times X$ is open in $X \times X$ if and only if $X$ is a discrete space.

Proof:

Let $x \in X$ then $(x,x) \in \triangle$, then by definition of product topology we can find a basic open set $U \times V$ such that $(x,x) \in U \times V \subseteq \triangle$. Then $U \times V = \{(x,x)\}$ so $\{(x,x)\}$ is an open set in $X \times X$. Now consider the projection map $p_{X}: X \times X \rightarrow X$, this is an open map and since $\{(x,x)\}$ is open in $X \times X$ it follows that $\{x\}$ is open in $X$, so $X$ is discrete.

Let us assume now that $X$ is discrete. Then if $(x,x) \in \triangle$ then since $\{x\}$ is open in $X$ then in particular: $(x,x) \in (\{x\} \times \{x\}) \subseteq \triangle$ so $\triangle$ is open. Is this OK?

Question 3. Show that if $A \cap \overline{B} = \emptyset = \overline{A} \cap B$ then $bd(A \cup B) = bd(A) \cup bd(B)$, here $bd$ denotes boundary.

Stuck with this one for a while. How to prove it?

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Aren't these three questions more or less unrelated? I mean: okay, they all are about topology, but I think it would be much better to ask them as separate questions. As for question 2: How do you justify "Then $U\times V = \{(x,x)\}$" I'm not quite clear how you get to that. The other direction is okay, of course. –  t.b. Aug 7 '11 at 8:01
@Theo Buehler: doesn't that imply that $U=\{x\}$? if $U \setminus \{x\} \neq \emptyset$ then pick $a$ in the set difference, then $(a,x) \in U \times V \subset \triangle$ so $a=x$, a contradiction. Is this wrong? –  user10 Aug 7 '11 at 8:09
No, this is right of course, but it looked like a somewhat unjustified conclusion. –  t.b. Aug 7 '11 at 8:10
Hint to start you on (3): If $x \in bd(A)$, either $x \in A$ (in which case $x \notin \overline{B}$, or $x \notin A \cup B$. Now think of what must be in a neighbourhood of $x$. –  Robert Israel Aug 7 '11 at 8:13
@Theo Buehler: Sorry to bother you, I already asked Brian about his proof but it seems he didn't read it and I really want to know how to solve this exercise. Can you please have a look at the last two lines of his proof? how do we know that $cl(B) \cap cl(A) \neq \emptyset$? we know that $x \in cl(A)$, how do we know that $x \in cl(B)$ as well? –  user10 Aug 9 '11 at 7:08

Your proof for (2) is basically fine, though the claim ‘Then $U \times V = {(x,x)}$’ could use a little more explicit justfication.

For (3), let $x \in \operatorname{bdry}(A \cup B)$; then $x \in \operatorname{cl}(A \cup B) = \operatorname{cl}A \cup \operatorname{cl}B$ and $x \in \operatorname{cl}\left(X \setminus (A \cup B)\right)$. Without loss of generality say $x \in \operatorname{cl}A$; I claim that $x \in \operatorname{bdry}A$. If not, then $x \notin \operatorname{cl}(X \setminus A)$, and there is an open set $V$ such that $x \in V$ and $V \cap (X \setminus A) = \varnothing$. But then $x \in V \subseteq A \subseteq A \cup B$, so $V$ is an open nbhd of $x$ that doesn’t intersect $X \setminus (A \cup B)$, and $x \notin \operatorname{cl}\left(X \setminus (A \cup B)\right)$. This contradiction establishes the claim, and it follows that $x \in \operatorname{bdry}(X \setminus A) \subseteq \operatorname{bdry}A \cup \operatorname{bdry}B$ and hence that $\operatorname{bdry}(A \cup B) \subseteq \operatorname{bdry}A \cup \operatorname{bdry}B$.

Now let $x \in \operatorname{bdry}A \cup \operatorname{bdry}B$; we may assume that $x \in \operatorname{bdry}A$, so that $x \in \operatorname{cl}A$ and $x \in \operatorname{cl}(X \setminus A)$. Certainly we then have $x \in \operatorname{cl}A \subseteq \operatorname{cl}(A \cup B)$, so to show that $x \in \operatorname{bdry}(A \cup B)$, we need only show that $x \in \operatorname{cl}\left(X \setminus (A \cup B)\right)$. By hypothesis $A \cap \operatorname{cl}B = B \cap \operatorname{cl}A = \varnothing$, so $\operatorname{cl}B \subseteq X \setminus A$ and $\operatorname{cl}A \subseteq X \setminus B$. But then $\operatorname{cl}\left(X \setminus (A \cup B)\right) = \operatorname{cl}\left((X \setminus A) \cap (X \setminus B) \right) \supseteq \operatorname{cl}\left(\operatorname{cl}B \cap \operatorname{cl}A \right) =$ $\operatorname{cl}B \cap \operatorname{cl}A$. Certainly $x \in \operatorname{cl}A$, so we’re done if $x \in \operatorname{cl}B$.

Edit:

If $x \notin \operatorname{cl}B$, let $V = X \setminus \operatorname{cl}B$, so that $V$ is an open nbhd of $x$ contained in $X \setminus B$. Let $W$ be any open nbhd of $x$; then \begin{align*}W \cap \left(X \setminus (A \cup B) \right) &= W \cap (X \setminus A) \cap (X \setminus B)\\ &\supseteq W \cap (X \setminus A) \cap V\\ &= (V \cap W) \cap (X \setminus A)\\ &\ne \varnothing, \end{align*} since $x \in \operatorname{cl}(X \setminus A)$, and hence $x \in \operatorname{cl}\left(X \setminus(A \cup B)\right)$ anyway.

To see what’s really going on, notice that in principle $\operatorname{bdry}(A \cup B)$ can fail to contain $\operatorname{bdry}A$. This situation can already be illustrated in the real line, with $A = [0,1)$ and $B = [1,2]$, for instance. Here $1 \in \operatorname{bdry}A = \{0,1\}$, but $1 \notin \operatorname{bdry}(A \cup B) = \{0,2\}$. The hypothesis that $A \cap \operatorname{cl}B = B \cap \operatorname{cl}A = \varnothing$ rules out this kind of situation by ensuring that a point in $\operatorname{bdry}A$ cannot already belong to $B$ and so cannot end up in $\operatorname{int}(A \cup B)$ (and similarly with $A$ and $B$ interchanged).

(There are various ways to define the boundary of a set $A$; here I’ve used the definition $\operatorname{bdry}A = \operatorname{cl}A \cap \operatorname{cl}(X \setminus A)$.)

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thanks as a lot, as usual your answers are very helpful! –  user10 Aug 7 '11 at 10:12
one question: in the last line, how do we know that $x \in cl(B)$? we know that $x \in cl(A)$, why is $x \in cl(B) \cap cl(A)$? –  user10 Aug 8 '11 at 23:18
@user10: At this point I’m not sure whether I slipped up and misread $\cap$ as $\cup$, or whether I simply left out some steps. At any rate, I’ve now fixed it. –  Brian M. Scott Aug 9 '11 at 8:13

By definition of product topology we can find a basic open set U×V such that $(x,x)\in U\times V\subseteq \Delta$. Then $U\times V=\{(x,x)\}$.

How did you get that if there is a basis containing $(x,x)$, then the basis is $U\times V=\{(x,x)\}$?

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Because it's a subset of the diagonal.... If $U \neq \{x \}$ or $V \neq \{x \}$, then we have some $(a,x) \in U \times V$ where $a \neq x$ or some $(x,b) \in U \times V$ where $b \neq x$, and in either case that point is not in $\Delta$, contradiction. –  Henno Brandsma Feb 7 '12 at 21:15