[Math] Supremum of a set of rationals

proof-verificationproof-writingreal-analysissupremum-and-infimum

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.

Best Answer

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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