Proof than an unbounded interval = $\mathbb{R}$

Question

My approach:

Let $\varnothing \neq A \subseteq \mathbb{R}$.

$A$ is unbounded, therefore $\forall x \in A$ , $\exists\ M_1,M_2 \in \mathbb{R}$

such that $M_1\leq x \leq M_2$, therefore $A \subseteq [-\infty, \infty ]$

In the same manner, $A$ is unbounded therefore $\forall x \in A$ , $\exists\ M_1,M_2 \in \mathbb{R}$

such that $M_1 < x < M_2$, therefore $\exists\ a_1,a_2 \in A$ such that $a_1 < x < a_2$ , (otherwise A would be bounded) and therefore $A \subseteq (-\infty, \infty )$.

Is this a legit proof? what should I improve \ change?

Answer 1

There is a lot to clean up in your proof. In your first line, you should note that you assume that A is an unbounded interval (and that unbounded means it is not bounded from above or from below).

Your conclusion does not prove the proposition, since you have merely shown that if $x \in A$ then $x \in R$, which is trivial.

Here's a sketch of how to write prove it in a standard way. The idea of the proof is to let $x \in \mathbb{R}$ be arbitrary, and then show that it is in A, proving that $A = \mathbb{R}$.

Let $x \in R$ be arbitrary. Assume for contradiction that $x \notin A$. Let $y \in A$, since $A \ne \emptyset$, and therefore that $x \ne y$. If $y<x$, then $z<x$ for all $z \in A$ (why?), so $A$ is bounded from above, contradiction! If $y>x$, then similarly, $z>x$ for all $z \in A$, so $A$ is bounded from below, contradiction.

So since our assumption let to a contradiction, the assumption ($x \ne A$) is wrong, so $x \in A$

Answer 2

You haven't properly written what being unbounded means, or properly proved that $\forall x\in A\; \exists \,a_1,a_2\in A : a_1<x<a_2$, and even if you did, that wouldn't be enough. You need to prove it $\forall x\in\mathbb{R}$, and then use the fact that $A$ is an interval.

'$A$ is not bounded from above' means that $\neg \exists x\in\mathbb{R} \,\forall a\in A: a\le x$, which is equivalent to $\forall x \in \mathbb{R} \;\exists a\in A : x<a$.

'$A$ is not bounded from below' means that $\neg \exists x\in\mathbb{R} \,\forall a\in A: x\le a$, which is equivalent to $\forall x \in \mathbb{R} \;\exists a\in A :a<x$.

'$A$ is unbounded' means that it have both of these properties, that is $\forall x \in \mathbb{R} \;\exists a_1,a_2\in A :a_1<x<a_2$

'$A$ is an interval' means that $\forall a_1,a_2\in A \forall x\in \mathbb{R}: \big((a_1<x<a_2)\Rightarrow (x\in A)\big)$

From putting these two conditions together we can get that $\forall x\in\mathbb{R}:x\in A$, that is $\mathbb{R}\subset A$. Oobviously, we have also $A \subset \mathbb{R}$, which means that $A=\mathbb{R}$

Answer 3

Let me go over your approach, and then suggest some improvements. It seems that you're trying to prove the following result:

• If $A\subseteq\Bbb R$ is an interval with no upper bound or lower bound, then $A=\Bbb R.$

Please let me know if that is not the case. Now, let's get into your approach.

Let $\varnothing \neq A \subseteq \mathbb{R}$.

If I'm correct about the result you're trying to prove, none of this needs to be said. We already know that $A\subseteq\Bbb R,$ and since $\varnothing$ is bounded, then we can conclude that $A\ne\varnothing.$

$A$ is unbounded, therefore $\forall x \in A$ , $\exists\ M_1,M_2 \in \mathbb{R}$ such that $M_1\leq x \leq M_2$,

This doesn't make much sense. For any non-empty subset $B$ of $\Bbb R$ and any $x\in B,$ there exist $M_1,M_2$ in $\Bbb R$ such that $M_1\le x\le M_2.$ In particular, $x\in \Bbb R$ and $x\le x\le x.$

therefore $x \subseteq [-\infty, \infty ]$

First of all, what is $[-\infty,\infty]$? Second of all, since when is $x$ a set? It was supposed to be a real number in $A,$ wasn't it? I suspect you meant to say "$x\in[-\infty,\infty],$" but even then, we have to address the first issue.

In the same manner, $A$ is unbounded therefore $\forall x \in A$ , $\exists\ M_1,M_2 \in \mathbb{R}$ such that $M_2 < x < M_2$,

Again, this has nothing whatever to do with $A.$ Also, I suspect that you meant "$M_1<x<M_2$," instead, since we cannot have $M_2<x<M_2.$ For any non-empty subset $B$ of $\Bbb R$ and any $x\in\Bbb R,$ there exist $M_1,M_2\in\Bbb R$ such that $M_1<x<M_2,$ because $\Bbb R$ is totally ordered, but has no least element or greatest element. In particular, $x-1,x+1\in\Bbb R$ and $x-1<x<x+1.$

therefore $\exists\ a_1,a_2 \in A$ such that $a_1 < x < a_2$, (otherwise A would be bounded)

True!

and therefore $x \subseteq (-\infty, \infty )$.

Again, when did $x$ become a set? I suspect you meant to say "$x\in(-\infty,\infty),$" which is true, but we didn't have to prove this. This is simply saying that $x\in\Bbb R,$ which we already knew, since $x\in A$ and $A\subseteq\Bbb R.$

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In order to show that $A=\Bbb R,$ knowing that $A\subseteq\Bbb R$ already, we must show that $A\supseteq\Bbb R.$ To do that, we must show that for any $x\in\Bbb R,$ we have $x\in A.$

We will need to use the three facts we know about $A$ in order to do this:

• $A$ is an interval: that is, for all $a_1,a_2\in A$, if $x\in\Bbb R$ with $a_1<x<a_2,$ then $x\in A.$
• $A$ has no lower bound: that is, for all $x\in\Bbb R,$ there is some $a_1\in A$ with $a_1<x.$
• $A$ has no upper bound: that is, for all $x\in\Bbb R,$ there is some $a_2\in A$ with $x<a_2.$

Can you see how to put the three pieces together?