0
$\begingroup$

I'm having trouble with this. Suppose that $U_1, U_2, \dots U_n$ are closed subsets of $\mathbb{R}^n$. I know that each $U_i' \subseteq U_i$ where $U_i'$ is the derived set but this doesn't feel like enough to show their cartesian product is closed.

Edit:

Suppose $S = U_1 \times U_2 \times \dots \times U_n$. Let $x \in \bar{S}$. Then $x \in \bar{U_i} = U_i$. Does this imply $x \in S$? I don't see why?

$\endgroup$
3
  • $\begingroup$ Prove $\overline{\prod A_i}=\prod \overline{A_i}$ for any $A_i$ $\endgroup$
    – Tanius
    Commented Nov 15, 2015 at 18:35
  • $\begingroup$ @TRmate is my edit in the ballpark of what you are suggesting? $\endgroup$
    – Chris
    Commented Nov 15, 2015 at 23:40
  • $\begingroup$ @Chris I have edited my answer to be more detailed. Haven't made the general case more detailed though, but I think you ought to be able to do that yourself. $\endgroup$
    – MickG
    Commented Nov 16, 2015 at 6:59

2 Answers 2

3
$\begingroup$

The projections $p_i$ onto the $i$-th coordinate are continuous.

$U_1 \times U_2 \times \ldots \times U_n = \cap_{i=1}^n (p_i)^{-1}[U_i]$ which is an intersection of closed sets so closed.

$\endgroup$
1
  • $\begingroup$ Thanks! But I have not worked on continuous projections yet. $\endgroup$
    – Chris
    Commented Nov 15, 2015 at 23:39
0
$\begingroup$

Henno's answer offers a neat solution.

You can also prove that the derived set of the product is contained in it. That equates to the product being closed. So take a limit point of the product. This means there is a sequence (OK, a net, but in our case $\mathbb{R}^n$ is a metric space so sequences suffice) in the product converging to it. But convergence is equivalent to componentwise convergence, so the sequence in question yields $n$ sequences in $U_i$ converging to the $i$th component of the limit of the "big" sequence. The $U_i$ are closed, meaning those sequences' limits are in the $U_i$s. But the limit of the "big" sequence is the "product" of those limits, and therefore belongs to the product of the $U_i$s.

A similar argument shows, as TRmate suggested in his comment, that the closure of the product is (well, contained in -- but one can modify the argument for the other inclusion) the product of the closures, no matter what the sets that are being "multiplicated" are.

Using nets, this can probably be generalized to subsets of arbitrary topological spaces.

Update

I do not quite understand your edit. What is $U_1,U_2,\dotsc,U_n$ supposed to mean? If it is the cartesian product, it should be $U_1\times U_2\times\dotso\times U_n$, with $\times$, not commas.

In this hypothesis, let $x\in\overline S$, with $S=U_1\times\dotso\times U_n$. What does $x\in\overline S$ mean? You should know $\overline S=S\cup S'$, with $S'$ the derived set. If you do not, tell me and I will add a proof. Now if $x\in S\cup S'$, then it can be that $x\in S$, or that $x\in S'$, or both. If $x\in S$, we are done. If $x\in S'$, we need to show $x\in S$ as well. But if $x\in S'$, since $S$ is a subset of the metric space $\mathbb{R}^n$, you ought to know that, $x$ being a limit point, one can find $x_k\to x$ such that $x_k\in S$ for all $k$. Again, if you do not know this, I can add a proof. Denote $x_k^{(i)}$ the $i$th component of $x_k$. $x_k$ has $n$ components since it is in $S=U_1\times\dotso\times U_n$. You ought to know $x_k\to x\iff x_k^{(i)}\to x^{(i)}$ for all $1\leq i\leq n$. Again, I can add a proof if you need one. So what have we just done? We have proved $x^{(i)}\in U_i'$ for all $i$. But $U_i'\subseteq U_i$, as you state you know, so $x^{(i)}\in U_i$ for all $i$. But then $x=(x^{(1)},\dotsc,x^{(n)})\in U_1\times\dotso\times U_n=S$, by definition of cartesian product. So $S'\subseteq S$, hence $S$ is closed.

Is it clear now?

To expand on Henno's solution, if you have $U_i\subseteq X$ for all $i$ and consider $U_1\times\dotso\times U_n$, you can easily see:

$$U_1\times\dotso\times U_n=(U_1\times X\times\dotso\times X)\cap\dotso\cap(X\times\dotso\times X\times U_n)=\bigcap_{i=1}^nX\times\dotso\times U_i\times\dotso\times X.$$

But if $\pi_i(x_1,\dotsc,x_n):=x_i$, then those sets being intersected are $\pi_i^{-1}(U_i)$, and since the product topology is by definition the coarsest topology making $\pi_i$ continuous for all $i$, we know $\pi_i^{-1}(U_i)$ is closed for all $I$, and the intersection of closed sets is closed.

Update 2

From the comment discussion continued in chat: 1, 2, image in previous message, 3, 4.

Update 3

After this bit, the chat was truncated by his disappearance. Perhaps commenting on this answer will attract his attention again…

$\endgroup$
15
  • $\begingroup$ I've tried a path in my edit but I don't think it will lead me where I want to go? $\endgroup$
    – Chris
    Commented Nov 15, 2015 at 23:40
  • $\begingroup$ Basically my argument proves $(U_1\times\dotso\times U_n)'\subseteq\prod U_i'$. $\endgroup$
    – MickG
    Commented Nov 16, 2015 at 7:11
  • $\begingroup$ That $x$ may be in a product of some $U_i$'s and some $U_i'$'s, so the sequence might have constant components. So scratch my last comment out :). $\endgroup$
    – MickG
    Commented Nov 16, 2015 at 7:14
  • $\begingroup$ Your answer makes sense but as I have not studied series (only accumulation points and basic topology) the notation is a bit foreign. From what I gather you are arguing that any of the limit points will converge to be in S. And yes my edit had an error that I have fixed. $\endgroup$
    – Chris
    Commented Nov 16, 2015 at 13:59
  • $\begingroup$ Limit points do not converge. Let us try some dialogue. I wish I could say: let us move to chat but that would impede MathJax rendering and might make our conversarion impossible. So... What is a limit point to you? $\endgroup$
    – MickG
    Commented Nov 16, 2015 at 14:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .