Cartesian product of open intervals is open (and of closed intervals is closed) [closed]

I have encountered a problem regarding basic topology and am not sure how to approach the proof.

The theorem states:

For $a_1,b_1,...,a_N,b_N\in\mathbb{R}$, s.t. $a_i<b_i$ for $i=1,..,n$, show that $(a_1,b_1)\times\cdots\times(a_N,b_N)$ is open and $[a_1,b_1]\times\cdots\times[a_N,b_N]$ is closed in $\mathbb{R}^N$.

I am not too familiar with topology, and am not really sure how to approach either case. Any recommendations?

closed as off-topic by user21820, Namaste, Simply Beautiful Art, ThePortakal, Paramanand SinghOct 1 '17 at 16:01

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How you should approach this problem depends on your background, how much you've already learned, and what you're expected to use.

I would presume that the working definition of an open set in $\mathbb{R}^n$ is the following:

$X\subseteq\mathbb{R}^n$ is open if for any point $x\in X$ there exists an open ball $B(x,r)$ such that $B(x,r)\subseteq X$. Here for any $x\in X$ and $r>0$, $B(x,r)=\{y\in X\mid d(x,y)<r\}$, where $d(x,y)$ is the usual Euclidean distance.

Note that for $\mathbb{R}$ this boils down to

$X\subseteq\mathbb{R}$ is open if for any point $x\in X$ there exists a number $r>0$ such that $(x-r,x+r)\subseteq X$.

So, let's start with the open part of the problem. For brevity, let's denote $X=(a_1,b_1)\times\cdots\times(a_N,b_N)\subset\mathbb{R}^n$. Pick any point $x\in X$. Then $x=(x_1,\ldots,x_N)$, where $x_1\in(a_1,b_1)$, …, $x_N\in(a_N,b_N)$. We know that open intervals are, well, open, so:

• since $x_1\in(a_1,b_1)$, there exists $r_1>0$ such that $(x_1-r_1,x_1+r_1)\subseteq(a_1,b_1)$;
• since $x_N\in(a_N,b_N)$, there exists $r_N>0$ such that $(x_N-r_N,x_N+r_N)\subseteq(a_N,b_N)$.

Now let $r=\min\{r_1,\ldots,r_N\}$. Note that for each $i=1,\ldots,N$, we have $(x_i-r,x_i+r)\subseteq(x_i-r_i,x_i+r_i)\subseteq(a_i,b_i)$. Using that, it only remains to show that $B(x,r)\subseteq X$, which I'll leave to you as an exercise.

So the product topology on $\Pi_{\alpha \in \mathcal A}X_{\alpha}$ is defined as the topology generated by subbase $\cup_{\alpha \in \mathcal A}\mathcal C_{\alpha}$ (i.e. the coasest topology for which all the projections $\{\pi_{\alpha}\}$ are continuous), where $$\mathcal C_{\alpha} = \{\pi_{\alpha}^{-1}(A): \, A \in \mathcal T_{\alpha}\}$$

And in your case, $\mathcal A=\{1, 2, 3, \ldots N\}$, and $X_{\alpha}=\mathbb R$

Let's take $O_i = \mathbb R_1 \times \mathbb R_2 \times \ldots \times (a_i, b_i) \times \ldots \times \mathbb R_N$, and by the definition of product topology, $O_i$ is open, because $O_i = \pi_i^{-1}((a_i,b_i))$, where $(a_i, b_i) \in \mathcal T(\mathbb R)$, the usual topology for $\mathbb R$, which means $O_i$ belongs to the subbase, and thus belongs to the topology of Cartesian space.

Let $O = \cap_{i=1}^{N} O_i =(a_1,b_1)\times (a_2, b_2)\times \ldots \times (a_i, b_i)\times \ldots \times (a_N, b_N)$. Since finite union of open sets are still open by the definition of topology, we conclude $O$ is open. We are done.

Similar for closed sets.

• Hm, I am not really sure I understand what the product topology, subbase or "projections are continuous" means. We haven't gone over a lot of topology and only know the basics. Is there anyway to simplify the proof? – The math god Sep 25 '17 at 3:13
• @Themathgod Ah, I mentioned that because you labeled it as general topology. I think you could then try to read my $O_i$ and $O$, could you see that $O_i$ is open? – Yujie Zha Sep 25 '17 at 3:14
• I am a bit confused about $O_i$ being open. Any chance you could elaborate? My apologies, just started topology a couple days ago. – The math god Sep 25 '17 at 3:24
• @Themathgod I think there two ways to see it, depending on what you've learned. If you think $\mathbb R^N$ is a metric space with Euclidean metric on it, you could check that $O_i$ is open by checking for any point $x \in O_i$, there exists a real number ε > 0 such that, given any point $y$ with $d(x, y) < \epsilon$, $y$ also belongs to $O_i$. (I suspect that this is what you want) – Yujie Zha Sep 25 '17 at 3:44
• @Themathgod Or, if you could go with the general topology way to get $O_i$ is open from the definition of the product topology on your product space. To explain that a bit more: if we project the set $O_i$ onto $i$th axis, you'll get $(a_i,b_i)$,and that′s what I meant for the project function $\pi_i$. And$(a_i, b_i)$ is open, so $O_i$ is open too - this is how product topology is define, i.e. how one type of open sets on the Cartesian product space is defined. – Yujie Zha Sep 25 '17 at 3:57