Supremum of a set of rationals

Question

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.

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

in your first part, showing b is an upper bound, your proof is valid. then to show that b is the supremum, proof by contradiction can be as follows:

let c $\in \Re$ be an upper bound of T, assume c<b.

since $\mathbb{Q}$ is dense in $\Re$, there exist q $\in \mathbb{Q}$ such that q $\in (c,b) \Rightarrow$ c<q<b. then q $\in$ T. then c is not an upperbound, hence a contradiction is reached. therefore b is the supremum.

the same approach also works for proofing supremum for an open interval T=$\mathbb{Q} \cap (a,b)$

note that provided b $\in \Re$, b can be an irrational number.