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I am preparing for my exam right now and therefore practicing by doing some exercises. I need help for the following task:

Let $A\subset\Bbb R^m$ and $B\subset\Bbb R^n$ be closed subsets. Then $A \times B$ is a closed subset of $\mathbb{R^{m+n}}$

I already found a good answer for a similar task, where $A\subset \mathbb{R}$ and $B\subset\mathbb{R}$ here.

My problem is, that I want to solve this task in a similar way. I want to show that $(A\times B)^c$ is open by doing all steps @DanielFischer did... but the thing here is, that we are working with $\mathbb{R^m}$ and $\mathbb{R^n}$ with m,n also being $\geq2$.

If we have $a\in A\subset \Bbb R$ and $b\in B\subset \Bbb R$, we can get the open balls $B(a,\varepsilon_1)$ and $B(b,\varepsilon_2)$ and then get $B((a,b),\min(\varepsilon_1,\varepsilon_2))\subset B(a,\varepsilon_1)\times B(b,\varepsilon_2)$. And then we want to prove that $(x,y)\in B((a,b),\min(\varepsilon_1,\varepsilon_2))$.

But I don't see how I could write this with $\mathrm a=(a_1,\mathrm,a_n)$ and $\mathrm b=(b_1,\ldots,b_n)$ being vectors. I mean $a(\in \Bbb R)\times b(\in \Bbb R)$ is obviously $(a,b)$. But what is $(a_1,\ldots,a_n)\times (b_1,\ldots,b_n)$

I don't know if its a dumb question but I hope someone of you can help me out.

Also...I found another solution that seems pretty easy but I can't quite follow it:Is product of two closed sets closed?

"$A$ is closed in $X$, so $A^c$ is open, likewise for $B$ in $Y$.

Moreover, $A^c \times Y$ and $X \times B^c$ are both open in $X \times Y$.

Thus $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c) $$ is open. Hence $A \times B$ is closed."

The problem here is that I can't follow why ($A^c \times Y$) and ($B^c\times X)$ is supposed to be open. Also why is ($A^c \times Y$)$\cup$ ($B^c\times X)$ open? We only learned that the intersection of two opened sets is definetly open. Is there like any proof for my problem?

I am thankful for any advice.

PinkyWay
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Analysis_Mark
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2 Answers2

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Alternative proof using sequences: Let $((a_k, b_k))_k$ be a sequence in $A \times B$ that converges to $(a,b) \in \mathbb{R}^{m}\times\mathbb{R}^n$. Then $\|a_k -a\| \leq \| (a_k,b_k) - (a,b) \| \to 0$ and thus $\| a_k - a \| \to 0$. Since $(a_k)_k$ is a sequence in $A$ and $A$ is closed, we find $a \in A$. We can apply the same argument and find $b \in B$. In summary: $(a,b) \in A \times B$.

Neckverse Herdman
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    Thx for your anser...But why is $||a_n-a||\leq ||(a_n,b_n)-(a,b)||$ if I may ask? – Analysis_Mark May 18 '22 at 19:21
  • Try to apply the definition of the 2-norm for both and it should be fairly obvious ... if not, definitely let me know! – Neckverse Herdman May 19 '22 at 01:26
  • Well its not that obvious for me. I know that $||a_k-a||=\sqrt{(a_{1,k}-a_1)^2+...+(a_{n,k}-a_n)^2}$. But how does it look like for $||(a_k,b_k)-(a,b)||$. We have $\sqrt{((a_{1,k},b_{1,k})-(a_1,b_1))^2+...+((a_{n,k},b_{n,k})-(a_n,b_n))^2}$. The only I idea I have why the second should be bigger is that probably $((a_{i,k},b_{i,k})-(a_i,b_i))^2\geq (a_{i,k}-a_i)$ $\forall i=1,...,n$. But why is that so. Is there like a good proof for that? – Analysis_Mark May 19 '22 at 11:29
  • I changed the $n$'s into $k$'s in my post, excuse me for the confusion. Your first statement is correct, but your second doesn't make much sense to me, since it seems like you are squaring vectors. (I'll write the component indices of a vector as a superscript.) Remember that $| (a_k, b_k) - (a,b) | = |(a_k - a, b_k -b)| = | (a_k^{(1)} - a^{(1)}, \ldots, a_k^{(m)} - a^{(m)}, b_k^{(1)} - b^{(1)}, \ldots, b_k^{(n)} - b^{(n)}) | = \sqrt{(a_k^{(1)} -a^{(1)})^2 + \ldots + (a_k^{(m)} - a^{(m)})^2 +: : \text{something positive}}$. – Neckverse Herdman May 19 '22 at 11:45
  • But why are we allowed to do that. I mean If we would have the sequence $x_k=(a_{1,k},...,a_{n,k},b_{1,k}...,b_{n,k})$ and the vector $x=(a_1,...,a_n, b_1,..., b_n)$ I would totally understand that $||x_k-x||=||(a_{1,k}-a_1, ..., a_{n,k}-a_n, b_{1,k}-b_1,..., b_{n,k}-b_n)||=\sqrt{(a_{1,k}-a_1)^2+...+(a_{n,k}-a_n)^2+(b_{1,k}-b)^2+...+(b_{n,k}-b_n)^2}$. But here we have (a_k-a,b_k-b) being a pair of two vectors.We can't just split the pairs up or am I wrong...Maybe there is a rule for that and I just don't know it. If you have a link or something, I would be very grateful if you could send it. – Analysis_Mark May 19 '22 at 12:15
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    But that is exactly the sequence and the vector that we have! (but you have to change the $a_{n,k}$ into $a_{m,k}$ because $a_k \in R^m$). I understand your confusion if this is the first time you're encountering this, but yes, we always just understand $(a,b) = ((a_1, \ldots, a_m),(b_1, \ldots, b_n))$ to equal $(a_1, \ldots, a_m, b_1, \ldots, b_n)$. Just like you implied (implicitly) in your post that $R^m \times R^n = R^{m +n}$, and $(a,b)$ is still an element of $R^{m +n}$! – Neckverse Herdman May 19 '22 at 13:02
  • Let us continue this discussion in chat. – Analysis_Mark May 19 '22 at 19:15
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We know that $A^c$ is open in $X$ and $B^c$ is open in $Y$. Let's just show that $A^c\times Y$ is open in $X\times Y$: Let $(x,y)\in A^c\times Y$. Since $A^c$ is open in $X$, let $U$ be an open neighborhood of $x$ such that $U\subseteq A^c$. Then, by the definition of the product topology on $X\times Y$, $U\times Y$ is an open neghborhood of $(x,y)$ such that $U\times Y\subseteq A^c\times Y$, so $A^c\times Y$ is open in $X\times Y$.

Also, $(A^c\times Y)\cup(X\times B^c)$ is open because it is the union of two open sets. The fact that an arbitrary union of open sets is open is part of the definition of a topology $\tau$ on some set $S$.

closedrhombus
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  • Thanks for your answer. Yeah ok I understand it now, that the union of two open sets is open. But I still don't understand why by definition of the product topology on$ X\cross Y$, $U \cross Y$ is an open neighborhoof of (x,y) such that.... – Analysis_Mark May 19 '22 at 11:47
  • The definition of the product topology on $X\times Y$ is that $A\subseteq X\times Y$ is open if and only if $A$ can be written as the union of sets of the form $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. – closedrhombus May 19 '22 at 11:51
  • Oh thank you. I didn't know about this lemma though. Do you probably have a link for the proof of this lemma? – Analysis_Mark May 19 '22 at 12:28
  • This is actually the definition of the box topology, but for finite products, the box topology is the same as the product topology. Take a look at https://en.m.wikipedia.org/wiki/Box_topology – closedrhombus May 19 '22 at 12:44