Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What have you tried in order to prove these? –  Nate Eldredge May 22 '12 at 14:23

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.