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Let $S\subset\mathbb R$ be a non-empty bounded set and let $T$ be a non-empty subset of $S$.

Is $T$ bounded from below and does $\inf(S)\le \inf(T)$ hold?

And if $S$ satisfies $\sup(S)=\inf(S)$ imply, does that mean $S$ is a singleton set?

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Yes to both.

Since $\inf(S)$ is a lower-bound for the set $S$ and $T\subset S$, then $T$ is also bounded from below and in particular $\inf(S)$ is a lower bound of $T$. Since $\inf(T)$ is the biggest lower bound of $T$ and $\inf(S)$ is some, then $\inf(S)\leq \inf(T)$.

For the second question. Assume that there exists $x\in S$ such that $x\neq \sup(S)$. Now either $x<\sup(S)$ or $x>\sup(S)$. The latter is obviously impossible since $\sup(B)$ is an upper bound of $S$, so $x<\sup(S)$. Since $\sup(S)=\inf(S)$, then this means that $x<\inf(S)$, which is again impossible since $\inf(S)$ is a lower bound for $S$. So $\sup(S)=\inf(S)$ is the only element of $S$. Note that this did not use the "inf" and "sup" properties at all: in general if any lower bound is at the same time an upper bound of a non-empty set, then with similar proof we may show that the set is a singleton.

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