Subset of a set problem with least upper bound Suppose that $S$ and $T$ are (non-empty) sets in $\mathbb{R}$, such that $T$ is bounded from above and $S\subseteq T$. Prove that $S$ is bounded from above and that
$$
\sup(S) \leq \sup(T).
$$
I tried to use proof by contradiction. I tired to set $\sup(S)-\sup(T)=\varepsilon$ for some $\varepsilon>0.$ But then what you do. I really cannot figure out anything. I don't have much tools to manipulate the sets. Although it is intuitively clear its true.
 A: This is much easier than it seems. Since $T$ is bounded above, $\sup T$ exists. If $s \in S$, then $s \in T$ so that $s \leq \sup T$ for all $s \in S$. Therefore, $\sup T$ is an upper bound for $S$. Therefore, $\sup S$ exists and by the definition of supremum, it must be that $\sup S \leq \sup T$.
A: If $\alpha\in S$, then we can see that $ S\subseteq T \Rightarrow \alpha\in T$. Let $y$ be, such that $y=\sup(T)\geq x \;\;\forall x\in T.$ In particular $y\geq\alpha \;\;\forall\alpha\in S.$ Let $z$ be, such that $z=\sup(S) \geq x \;\; \forall x\in S.$ Then $y\geq z$. From here you can complete the proof by pointing out that $S$ is bounded from above...
A: You need to be a bit more careful with your proof by contradiction. You need to show that $S$ is bounded AND $\sup(S) \leq \sup(T)$. If you seek a contradiction from the get-go, this means you need to assume $S$ is not bounded (which means the supremum of $S$ does not exist) OR $\sup(S)>\sup(T)$. The proof would be much cleaner if you first show that $\sup(S)$ exists before beginning the proof by contradiction.
Let $\sup(T) = M$. By definition of supremum, for all $t \in T$ this means $t \leq M$. Since $S \subset T$ it is also true that $s \leq M$ for any $s \in S$ since $s \in T$. Thus $M$ is an upper bound for $S$, so $S$ has a supremum. Now that you have established that the supremum of $S$ exists, you could consider a proof by contradiction to show $\sup(S) \leq \sup(T)$. To begin, let's suppose towards contradiction that $\sup(S) = N > \sup(T) = M$. Now we can make some cases to complete the proof.
Case $1$: $N \in S$. Then $N \in T$. Since $M$ is the supremum of $T$ it must be that $N \leq M$, yet towards contradiction we had assumed $N>M$ so we clearly have a problem if $N \in S$.
Case $2$: $N \notin S$. This is a good exercise to practice, so I will leave it to you!
