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Prove that a closed rectangle in $\mathbb{R^n}$ is a closed set.

My trial:

Define the closed rectangle by: $A=\{a_{1} \leq x_{1} \leq b_{1},a_{2} \leq x_{2} \leq b_{2}, ...,a_{n} \leq x_{n} \leq b_{n} \}$

A closed set is defined as a set that contains all its cluster points.

Distinguish between 2 cases:

Case1: $x \in A$.

Then there exist an interval $[a,b]$ such that $a \leq x \leq b$ by the definition of a closed rectangle and then $A \cap [a,b]$\ $\{x\} \neq \phi$.

Is this justification correct?

Case2:$x \notin A.$

Then distinguish between 2 cases:

Case a:

$\inf|| x - A|| = 0$

Then in this case $x$ is a cluster point of A (either A is closed or open it does not matter) then there exist a sequence $c_{k} \in A$ such that $c_{k} \rightarrow x$ as $k \rightarrow \infty.$ But since $c_{k} \in A$ then $a_{i} \leq c_{k} \leq b_{i}.$ Then by squeeze theorem and since $a_{i}$ & $b_{i}$ are constants we have $a_{i} \leq x \leq b_{i}.$ which means that $x \in A$ which is a contradiction to our assumption..... but I do not know what should I conclude then? ...... may be this case can not exist i.e. the infimum can not be equal zero?

Case b:

$\inf||x - A|| \neq 0 $

Then put $ a = \inf||x - A|| ,$ then the interval $[x - a, x + a]$ does not contain other points of $A$ and hence $x$ is not a cluster point of $A$ in this case..... is my argument correct in this case?

Could anyone help me filling the gaps in my proof please?

EDIT 1:

Case1: $x \in A$.

Then there exist $1 \leq i \leq n$ such that $a_{i} \leq x_{i} \leq b_{i}$ by the definition of a closed rectangle and then $A \cap [a_{i},b_{i}]$\ $\{x_{i}\} \neq \phi \forall i$.

And hence $x_{i}$ is a cluster point of $A$ $\forall i$ and $x \in A$ ...... is my new edit correct?

EDIT 2:

Case b:

The following case should be deleted because it is not a cluster point.

$\inf||x - A|| \neq 0 $

Then put $ a = \inf||x - A|| ,$ then the interval $[x - a, x + a]$ does not contain other points of $A$ and hence $x$ is not a cluster point of $A$ in this case..... is my argument correct in this case?

Intuition
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    Don't you need only the case that $x$ is a cluster point? – Hagen von Eitzen Feb 25 '19 at 09:52
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    $A={x\in\mathbb{R}^n:a_1\leq x_1\leq b_1,...,a_n\leq x_n\leq b_n}\subseteq \mathbb{R}^n$ and thus $A\cap [a,b]=\emptyset$ ...and $a\leq x\leq b$ for $x\in A$ and $a,b\in\mathbb{R}$ makes no sense – Peter Melech Feb 25 '19 at 10:01
  • case 1. False when a = b and x cannot be in an interval when n > 1. – William Elliot Feb 25 '19 at 10:03
  • I could not understand your comment @HagenvonEitzen – Intuition Feb 25 '19 at 10:28
  • So could you please provide me with a rigorous answer @PeterMelech – Intuition Feb 25 '19 at 10:30
  • So could you please provide me with a rigorous answer @WilliamElliot – Intuition Feb 25 '19 at 10:30
  • why I need only the case $x$ is a cluster point @HagenvonEitzen ? – Intuition Mar 03 '19 at 10:57
  • @PeterMelech so is this the only mistake I have in my solution? – Intuition Mar 03 '19 at 10:59
  • I have edited my solution ..... could you check it please?@PeterMelech – Intuition Mar 03 '19 at 17:16
  • I have edited my solution ..... could you check it please? ....... I will also remove the case when $x \notin A$ and $ \inf \neq 0$ @HagenvonEitzen ....... I see that it is irrelevant as you said above – Intuition Mar 03 '19 at 17:18
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    What does "Then there exist an interval [a,b] such that a≤x≤b" mean? $x$ is n-tuple point in $\mathbb R^n$ and $\mathbb R^n$ doesn't have an order. Are you saying $a,b$ are n-tuple points in $\mathbb R^n$ and $a \le x \le b$ is shorthand for $a_i \le x \le b_i$? If so what is the point? the rectangle is that? Or do you mean $a,b\in\mathbb R$ and $a\le x\le b$ is shorthand for $a\le x_i \le b$. If so, how do you know. ANd any way... what's the point. What does this case show? – fleablood Mar 03 '19 at 17:26
  • I have just edited this case at the end of my post @fleablood – Intuition Mar 03 '19 at 17:28
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    I don't understand your Edit 1: at all. To begin with there isn't one $i$. That's true for all $i$. Second, $[a_i,b_i]\subset \mathbb R$ and $\mathbb R\cap \mathbb R^n = \emptyset$ so so $A\cap[a_i,b_i]=\emptyset$ and thirdly, what does any of that have to do with cluster points, and fourthly ao what if all points in A are cluster points (which they aren't; the border points are not cluster points)? that's that's neither required nor sufficient to be closed. – fleablood Mar 03 '19 at 17:50
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    You only need one case. 1: $x$ is a cluster point of $A\implies$ $x \in A$. That's the only thing that is relevant. If $x$ is not a cluster point of $A$ it does not matter if $x$ in $A$ or not. If $x$ is in $A$ it does not matter if it is or is not a cluster point. If 2: $x\not \in A\implies x \not \in A$ is another case but it is completely equivalent (contrapositive) to $1$ so you only have to choose one or the other to prove. (Advice: prove 1. – fleablood Mar 03 '19 at 17:52
  • your comments are so valuable :) thank you ........ that was exactly what I am looking for ...... correction to my solution ........ could you please write an explicit answer for me so that I can accept an answer for this question?@fleablood – Intuition Mar 04 '19 at 03:00
  • @fleablood ..... do you mean that I need not edit case 1 and it was written correctly or what shall I do to correct it? – Intuition Mar 04 '19 at 03:03
  • why $\mathbb{R} \cap \mathbb{R^n} = \emptyset $? does not $(x, 0 ,0 .....,0)$ n-tuple belongs to $\mathbb{R}$ and $\mathbb{R^n}$? – Intuition Mar 04 '19 at 03:06
  • for your very first comment ..... so what is the correct way to write case 1?@fleablood I got confused...... actually this case is saying that the set A contains all the cluster points of A that lies inside A. – Intuition Mar 04 '19 at 03:44
  • For your second comment so how can I correct this case ? or what shall I do? @fleablood – Intuition Mar 04 '19 at 03:52
  • I think that for your last comment the conclusion of the implication is incorrect @fleablood – Intuition Mar 04 '19 at 04:03

2 Answers2

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I cannot comment yet so I must post in answer. Sorry about that.

Well any $[a_i,b_i]$ is closed in $\mathbb{R}$. Now the Cartesian product $A \times B$ of two closed set $A$ (closed in $X$) and B (closed in $Y$) is closed in $X \times Y$ (see Is product of two closed sets closed?). So then your rectangle is closed in $\mathbb{R}^n$ as the Cartesian product of $n$ closed sets in $\mathbb{R}$.

Plussoyeur
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    I'd almost call this an answer, except that it's just a little too topological for the given definition of closed. – Theo Bendit Feb 25 '19 at 10:03
  • @TheoBendit yes I agree. That's why I wanted to make a comment below the post instead of posting it as answer since the author seems to want a proof via cluster points. But I did not have enough reputation. Thanks to the upvote, I can now comment. – Plussoyeur Feb 25 '19 at 10:06
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    @Plussoyeur Another more ;) – Dog_69 Feb 25 '19 at 12:02
  • thank you for your solution but I need a correction to my solution not a new solution – Intuition Mar 03 '19 at 16:54
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I'll try though I prefer Plussoyeur's answer. Consider $x\in\mathbb{R}^n$ so that for every neighborhood $U\in \mathcal{U}(x)$, ( the system of neighborhoods) You have $$U\cap A\neq\emptyset.$$ (So $x$ is an clusterpoint of $A$.) Call $$P_i:\mathbb{R}^n\rightarrow\mathbb{R},\begin{pmatrix}x_1\\ \cdot\\\cdot\\x_i\\\cdot\\x_n\end{pmatrix}\mapsto x_i$$ the projection on the $i$'th component and for each $U\in\mathcal{U}(x)$ define $U_i=P_i(U)$ for $i=1,...,n$. Now by the product-topology the projections are open and thus $U_i$ is a neighborhood of $x_i$ that satisfies $$U_i\cap[a_i,b_i]\neq\emptyset$$ for $i=1,...,n$. Thus $x_i\in[a_i,b_i]$ for $i=1,...,n$ and so $x\in A$.

Peter Melech
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