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My question reads:

Let $a<b$ be real numbers and let the set $T$= $\mathbb{Q}$ $\cap$ $[a,b]$. Prove $sup T=b$.

My work so far:

(Showing $b$ is an upper bound). Let $y\in\ T$. Then $y\in\ [a,b]$ since the intersection of the rationals and an interval between two real numbers is just said interval. Then, $y\leq\ b$ which shows $b$ is an upper bound.

(Showing least upper bound-Proof by Contradiction) Let $w\in\ T$ and suppose $w< b$. Then by density, there exists an $x$ such that $b<x<w$.

Here is where I get stuck. I am not too sure how to continue with density here to arrive at a contradiction. I think I am on the right track, but I am missing a step somewhere here.

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  • $\begingroup$ I rephrased the title; "Proving a supremum" makes no sense. (You prove a statement, not a number.) $\endgroup$ – Andrés E. Caicedo Sep 17 '17 at 22:33
  • $\begingroup$ I think you are thinking too hard about $T=\mathbb Q \cap [a,b]$. It's easier to just think of: $T$ is all the set of all rational numbers $q$ so that $a \le q \le b$. The rest should be obvious. All such $q \le b$ so $b$ is upperbound. If $w < b$ then there is a $q$ so that $a\le \min(w,a) < q < b$ by the density of the rationals. So $w$ is not an upper bound. So $w$ is the least upper bound. $\endgroup$ – fleablood Sep 17 '17 at 23:26
  • $\begingroup$ You have an erroneously reversed inequality: If $b>w\in T$ then there exists an $x\in \mathbb Q$ with $w<x<b$ (NOT $b<x<w$).... So since $w\in T$ implies $w\geq a,$ you have $a<x<b$ and $x\in \mathbb Q$ so $x\in T.$... Hence $w\ne \sup T$ because $w <x\in T.$ $\endgroup$ – DanielWainfleet Sep 17 '17 at 23:43
  • $\begingroup$ @DanielWainfleet I was trying to do a proof by contradiction which is why I have the reversed inequality. $\endgroup$ – Sam Sep 18 '17 at 2:26
  • $\begingroup$ @DanielWainfleet I wanted to use a proof by contradiction for the last portion. $\endgroup$ – Sam Sep 19 '17 at 1:20
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proof-verification

(Showing $b$ is an upper bound of the set T). Let $y\in\ T$. Then $y\in\ [a,b]$ since the intersection of the rationals and an interval between two real numbers is just said interval. [I don't know what you are trying to say in this sentence. b is an upper bound of the set T simply because T is a subset of the interval [a,b]. ] Then, $y\leq\ b$ which shows $b$ is an upper bound of the set T.

(Showing least upper bound-Proof by Contradiction) Let $w\in\ T$ and suppose $w< b$. [This is not the correct way to prove by contradiction. Your argument breaks down from here.] Then by density, there exists an $x$ such that $b<x<w$. [This does not make sense after you assume that b is larger than w.]

Here is where I get stuck. I am not too sure how to continue with density here to arrive at a contradiction. I think I am on the right track, but I am missing a step somewhere here.


To fix the second step, assume instead, $b$ is not a strict upper of $T$, namely there exists a real number $c<b$ such that $c$ is an upper bound "of the set $T$" (one should not miss this phrase!). Now try to see what contradiction you can get.

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  • $\begingroup$ Would you mind taking a look at my improved proof if I post it here? $\endgroup$ – Sam Sep 19 '17 at 22:04
  • $\begingroup$ @Sam: you could update your post by adding your new proof under the old one. $\endgroup$ – Jack Sep 19 '17 at 22:09
  • $\begingroup$ Hopefully, it makes more sense now. $\endgroup$ – Sam Sep 19 '17 at 22:22
  • $\begingroup$ @sam: some phrases are redundant though, the updated proof looks good. $\endgroup$ – Jack Sep 19 '17 at 22:24
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    $\begingroup$ @sam: You are welcome. I think so. $\endgroup$ – Jack Sep 19 '17 at 22:29
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When you're showing that $b$ is an upper bound, $T\subset [a,b],$ so if $y\in T,$ then $y\in[a,b],$ and therefore $y\leq b.$ Your statement ("the intersection of the rationals and an interval between two real numbers is just said interval") is false, since for example $[\pi,2\pi]\cap\mathbb{Q}$ cannot contain the endpoints $\pi$ and $2\pi$.

When you are showing that $b$ is the least upper bound, you should not assume that $w\in T.$ For example, the supremum of the set $(0,1)$ is $1,$ but $1\not\in (0,1),$ so you should only be assuming that $w$ is an upper bound for $T$, and $w<b.$ As you said, by density, there is a $x\in\mathbb{Q}$ such that $w<x<b,$ since $(w,b)$ is an open interval and $\mathbb{Q}$ is dense. But since $x\in \mathbb{Q}\cap[a,b]=T,$ $w$ cannot be an upper bound for $T$. This contradiction proves that $b$ is in fact the least upper bound for $T$, which completes the proof.

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  • $\begingroup$ Okay, I see what I did wrong at the top. Then I cannot simply say $y\in\ [a,b]$ ? Also, I see what I did wrong in the second part of the proof by placing $w\in T$ . I think that is what I was trying to get to, but I did not how to say it correctly that I wanted to pick another upper bound for the set T. $\endgroup$ – Sam Sep 17 '17 at 22:31
  • $\begingroup$ Sure you can, I'm just saying that it follows because $A\cap B\subseteq A$ and $A\cap B\subseteq B$ are true for any sets $A$ and $B$, rather than any property of the real or rational numbers. $\endgroup$ – RideTheWavelet Sep 17 '17 at 22:32
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    $\begingroup$ You have to rely on the fact that given and $x < y;x,y \in mathbb R$ there will exists a $q\in \mathbb Q$ so that $x < q < y$. So if $w$ is anything so that $a < w < b$ then we can find a rational $q$ so that $a < w < q < b$. So $q \in T$ and $w$ is not an upper bound. So $b$ is the least upper bound. We do have to assume that given any two real $x<y$ we can find a rational $x < q < y$. But we can assume that. That is what is call Q being dense in R. $\endgroup$ – fleablood Sep 18 '17 at 2:35
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    $\begingroup$ To prove that $b$ is the least upper bound you have to prove that anything less is not an upper bound. If we pick $w \in T$ then that only proves that $w \in T$ is not an upper bound. It doesn't prove that everything else isn't an upper bound either. $\endgroup$ – fleablood Sep 18 '17 at 2:43
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    $\begingroup$ Right, to get $x\in T,$ we need $a\leq x\leq b,$ and $x\in \mathbb{Q}.$ What I'm saying is that it doesn't matter whether or not $w<a$ or $a\leq w,$ because in either case, we get to choose our $x$ to be a rational in the interval $(w,b)$ such that $a\leq x:$ when $a\leq w,$ choosing $x\in(w,b)$ guarantees $a\leq x,$ and when $w<a,$ we have $a<b,$ so choosing a rational $x\in(a,b)$ implies that $x\in(w,b)$, too. $\endgroup$ – RideTheWavelet Sep 18 '17 at 14:08
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I think you are thinking too hard about $T=\mathbb Q∩[a,b]$.

It's easier to just think of: $T$ is all the set of all rational numbers $q$ so that $a≤q≤b$.

The rest should now be obvious.

All such $q≤b$, so $b$ is an upper bound of $T$.

If $a< w<b$ (the density of the reals) then, (by the density of the rationals within the reals), there is a rational $q$ so that $a< w<q<b$.

So $w$ is not an upper bound of $T$.

So $b$ is the least upper bound of $T$.

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The Density of $\mathbb Q$ in $\mathbb R$. Lemma: for any $x,y \in \mathbb R$ such that $x < y$ then there exists a $q \in Q$ so that $x < q < y$.

Pf: Okay, I'm doing this off the top of my head. It's probably already been proven in your text. But...

Let $w = \frac {x+y}2 \in \mathbb R$. Then $x < w < y$. By the definition of the reals there is a sequence of rational $\{q_n\}\to w$. Thus or any $\epsilon > 0$ there are $q_i$ so that $|w -q_i|< \epsilon$. If we let $\epsilon < \frac {y-x}2$ then we have $x < w - \epsilon < w < w+ \epsilon < y$. And also have $w - \epsilon < q_i < w+ \epsilon$. So $ x < w -\epsilon < q_i < w+ \epsilon < y$. QED.

Okay, maybe that was overkill for me to prove, but it should be a basic fact you know.

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Addendum

Definition (I) of Least Upper Bound.

For a set $S$. $w$ is the least upper bound if i )for all $s\in S$ we have $s < w$. (i.e. $w$ is an upper bound.)

ii) if $v$ is an upper bound of $S$ then $w \le v$.

Definition (II) of Least Upper Bound.

For a set $S$. $w$ is the least upper bound if i )for all $s\in S$ we have $s < w$. (i.e. $w$ is an upper bound.)

ii) if $z < w$ then $z$ is not an upper bound.

Both of these definition are the same.

I $\implies$ II: If $w = \sup S$ and $z < w$. Then $z \not \ge w$ So $z$ is not upper bound.

II $\implies$ I: If $w = \sup S$ and $v$ is an upper bound then $v \not < w$ so $v\ge w$.

So if I were to do the proof using Definition I instead of definition II I'd say.

$b$ is an upper bound of $T$. Let $v$ be an upper bound $T$ as well. There is a rational $q$ so that $a < q < b$ so $q \in T$. So $a< q \le v$ and $v > a$. If $a < v < b$ then there are is a rational $r$ so that $a< v < r < b$. So $r$ is in $T$. So $v$ is not an upper bound with is a contradiction. So $v \ge b$. So $b$ is the least upper bound.

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  • $\begingroup$ I see what you are doing with "If a<w<b (the density of the reals) then, (by the density of the rationals within the reals), there is a rational q so that a<w<q<b" however, how does this give us b is a least upper bound? Is this not just showing w is not an upper bound? $\endgroup$ – Sam Sep 18 '17 at 2:38
  • $\begingroup$ The definition of least upper bound is that for any real number $w < b$ then $w$ isn't an upper bound. As $w$ was i) arbitrary and ii) not an upper bound, we know that any $w < b$ that $w$ will not be an upper bound. So that means $b$ is the least possible upper bound. $\endgroup$ – fleablood Sep 18 '17 at 2:41
  • $\begingroup$ Is the definition of least upper bound not if say a is an upper bound and b is an upper bound then $a\leq\ b$ means a is a least upper bound? $\endgroup$ – Sam Sep 18 '17 at 2:43
  • $\begingroup$ Yes, that is equivalent. I'll give an alternative. $\endgroup$ – fleablood Sep 18 '17 at 2:44
  • $\begingroup$ Okay, because I was trying to do my proof using contradiction to say b is a least upper bound. Then, is picking $w\in\ T$ okay to start with because I was given a comment that $w$ does not need to be in T. $\endgroup$ – Sam Sep 18 '17 at 2:47
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To show that $b$ is the least upper bound, suppose the contrary, that is, suppose there is another upper bound $w$ such that $w \lt b$ and try to get a contradiction.

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