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

I can show that $b\geq x$ for all $x\in T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.

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  • $\begingroup$ Take any other upper bound and show it must be greater than $b$ $\endgroup$ – Gregory Grant Aug 25 '15 at 18:13
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    $\begingroup$ Use the fact that between any two distinct reals is a rational number. $\endgroup$ – Akiva Weinberger Aug 25 '15 at 18:14
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Let $b=d_1.d_2d_3\dots d_k\dots$

Then the members of the sequence $b-0.\dots d_k\dots, b-0.0\dots d_k\dots, b-0.00\dots d_k\dots \dots$ are all rational, with $b-0.000\dots d_k\dots\to b$ as $k\to\infty$, hence the supremum of this sequence is $b$.

For example, let $b=\pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,\dots$. We have $b_n\in\mathbb{Q}, \lim_\limits{n\to\infty}b_n\to\pi\in\mathbb{I}$.

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Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $\mathbb{Q}$ there is some rational number $r$ such that $$M<r<b. \ \ \ (1)$$ Again by denseness of the rationals, $T$ is not empty. Take some $x \in T$. Then $a \leq x \leq M$. Combining this with the above yields $r \in T$, which gives $r \leq M$, a contradiction with $(1)$.

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