# Trying to show that $A\times \mathbb R^d$ is open or closed

$\def\R{\mathbb R}$Say there are two sets $A$ and $B$ where $A\subset \R^{m_1}$ and $B\subset \R^{m_2}$, then define $A\times B = \lbrace{c=(a_1,\ldots,a_{m_1},b_1,\ldots,b_{m_2})\in \R^{m_1+m_2}:a\in \R^{m_1},b\in \R^{m_2}\rbrace}$ . I am trying to show that there is a closed $X\subset \R^{m_1}$ such that $X\times \R^{m_2}$ is closed. I've started by saying there is an open $Y\subset \R^{m_1}$ such that $Y\times\R^{m_2}$ is open, but I'm not sure how to prove even that. Any advice to tackle this problem would be appreciated.

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You can use \in for $\in$, \times for $\times$ and \mathbb R for $\mathbb R$. – martini Sep 18 '12 at 18:55
Sure you mean "...there is a closed $X$ ..." instead of " ... for all closed $X$ ...", since the latter is true and the former is too easy, as $X = \emptyset$ will do. – martini Sep 18 '12 at 18:57
$X\times\Bbb R^{m_2}$ is closed for every closed $X\subseteq\Bbb R^{m_1}$ and open for every open subset $X\subseteq\Bbb R^{m_1}$. – Brian M. Scott Sep 18 '12 at 18:58
In the product topology, open $\times$ open is open and closed $\times$ closed is closed. Maybe you're not sure that the topology on $\mathbb R^{m_1+m_2}$ is the product topology of $\mathbb R^{m_1}$ and $\mathbb R^{m_2}$? – Hagen von Eitzen Sep 18 '12 at 19:28

I assume you're talking about $\mathbb{R}^m$ as a metric space with the Euclidean distance, as in the product topology your question follows from the definitions.

Take any open set $X \subset \mathbb{R}^{m_1}$. Since $X$ is open, for any point $p \in X$, there is an open ball $\mathrm{B}(p, \delta) \subset X$ centered at $p$ with radius $\delta$. We have $\mathrm{B}(p, \delta) \times \mathbb{R}^{m_2} \subset X \times \mathbb{R}^{m_2}$. Let $q \in \mathbb{R}^{m_2}$ and let $\mathrm{B}((p, q), \delta)$ be the open ball centered at $(p, q)$ with radius $\delta$. Clearly:

$$\mathrm{B}(p, \delta) \times \mathbb{R}^{m_2} \subset X \times \mathbb{R}^{m_2}$$

We claim that:

$$\mathrm{B}((p, q), \delta) \subset \mathrm{B}(p, \delta) \times \mathbb{R}^{m_2}$$

Let $x \in \mathrm{B}((p, q), \delta)$. Let $x_1$ be the point in $\mathbb{R}^{m_1}$ consisting of the first $m_1$ components of $x$. We have $d((p, q), x) < \delta$. By using the definition of the Euclidean distance, this forces $d(p, x_1) < \delta$ too. It follows that $x \in \mathrm{B}(p, \delta) \times \mathbb{R}^{m_2}$ as claimed.

Now we have:

$$\mathrm{B}((p, q), \delta) \subset \mathrm{B}(p, \delta) \times \mathbb{R}^{m_2} \subset X \times \mathbb{R}^{m_2}$$

Since $p$ and $q$ are arbitrary points, it follows that $X \times \mathbb{R}^{m_2}$ is open.

For the closed set, consider the open complement as you suggested.

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Just as a clarification: You are arguing that there will be a ball centered at (p,q) that will be a subset of X times R where delta is greater than zero. And this implies that X times R is open. – rioneye Sep 18 '12 at 19:52
@rioneye Yes, but it's not any $\delta$. It's the same $\delta$ that produces an open ball for $p$ in $X$. – Ayman Hourieh Sep 18 '12 at 19:53
Could you elaborate why $\mathrm{B}((p, q), \delta) \subset \mathrm{B}(p, \delta)\times R^{m_2}$? I may be having trouble saying that the $\delta$ delta must be the same. Is delta arbitrary such that we can say the delta is the same? Is there a way to say that $\mathrm{B}((p, q), \delta) = \mathrm{B}(p, \delta) \times \mathrm{B}(q, \delta)$ since that would make the solution more intuitive? – rioneye Sep 14 '13 at 19:02
@rioneye I added some more details to explain the inclusion you ask about. As for the equality $\mathrm{B}((p, q), \delta) = \mathrm{B}(p, \delta) \times \mathrm{B}(q, \delta)$, it's not true, so you cannot prove it or use it in this proof. You need an inclusion like the one I have in my answer. – Ayman Hourieh Sep 14 '13 at 20:54