Let $S$ and $T$ be nonempty bounded subsets of $\mathbb{R}$ with $S\subseteq T$. How do I prove that $\inf T\leq\inf S\leq\sup S\leq \sup T$?

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    $\begingroup$ Since $S \subset T$, every upper/lower bound of $T$ is also one of $S$. $\endgroup$ – Daniel Fischer Aug 26 '13 at 14:29
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    $\begingroup$ I changed \text{inf}T to \inf T, and similarly for \sup. This not only results in proper spacing before and after $\inf$ in expressions like $3\inf S$, but also affects positions of subscripts in expressions like $\displaystyle\inf_{x\in S}f(x)$ when those appear in a "displayed" setting rather than an "inline" setting. (In an "inline" setting, you'd see $\inf_{x\in S} f(x)$.) \inf and \sup are standard usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 26 '13 at 14:32

$S\subseteq T$ means every member of $S$ is a member of $T$.

$a = \inf S$ means $a$ is the largest lower bound of $S$, so it's $\le$ every member of $S$ and $\ge$ all other lower bounds of $S$.

If a number $b$ is $\le$ every member of $T$, and every member of $S$ is a member of $T$, then $b$ is $\le$ every member of $S$. Therefore $\inf T$ is $\le$ every member of $S$. Therefore $\inf T$ is a lower bound of $S$. Therefore $\inf T$ is $\le$ the largest lower bound of $S$. In other words $\inf T\le \inf S$.

A similar argument shows $\sup T\ge\sup S$.

The statement that $\inf S\le \sup S$ is true only if $S\ne\varnothing$. If $S\ne\varnothing$, then there exists $s\in S$. And we must then have $\inf S\le s\le\sup S$.

The fact that if $S\subseteq T$ then $\inf S\ge\inf T$ shows that $\inf\varnothing\ge$ all other "inf"s, so $\inf\varnothing=\infty$. Similarly $\sup\varnothing=-\infty$.



Direct proof:

  • Let $(x_n)_{n \in \mathbb{N}} \subseteq S$ be a sequence such that $\lim_{n \to \infty} x_n = \inf S$.
  • Since $S \subseteq T$ we have $(x_n)_{n \in \mathbb{N}} \subseteq T$ and so $\lim_{n \to \infty} x_n \geq \inf T$.

By contradiction:

  • Assume that $\inf S < \inf T$.
  • By definition of infimum, there exists $x \in S$ such that $x < \inf T$, but $x \in T$, contradiction with the definition of $\inf T$.

I hope this helps $\ddot\smile$

  • $\begingroup$ Mentioning sequences and limits seems to make this more complicated than necessary, since the concepts of inf and sup don't rely on the concept of limits. So your second argument is better than your first. $\endgroup$ – Michael Hardy Aug 26 '13 at 14:46
  • $\begingroup$ @MichaelHardy Yeah, but I think it emphasizes nicely the fact that the infimum or the supremum does not need to belong to the considered set. With standard spaces like $\mathbb{R}$ I find it more useful than the plain definition. $\endgroup$ – dtldarek Aug 26 '13 at 14:53

I went about the proof this way. Please tell me if I am missing something.

Case 1. If $A \subset B$ then there exists $b_1,b_2\in B$ such that $b_1,b_2 \notin A$. Let $a_1,a_2 \in A$ such that $a_1=$ inf A and $a_2$ = sup A. Suppose $b_1$=inf B and $b_2$=sup B than $b_1<a_1<a_2<b_2$ which implies inf $B<$ inf $A < $ sup $A<$ sup $B$

Case 2. If $A = B$ than every element in $A$ is in $B$. This implies that if $A$ is bounded above or below so is $B$ and vice versa. If the sup $B$ is defined to be the least upper bound and the inf $B$ is defined to be the greatest lowest bound. Than sup $B$ = sup $A$ and inf $B$ = inf $A$. \ 1 Since $A \subseteq B$ the following equality can be written as inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

  • $\begingroup$ I went about it this way as well, though there are more cases. It's possible for $\inf(T)=\inf(S)\leq\sup(S)<\sup(T)$ $\endgroup$ – David Farmilant Sep 18 '20 at 14:35
  • $\begingroup$ That is, S is the singleton $s$ which equals the $\inf(T)$ strictly less than $\sup(T)$. there are two other cases similar to this one which regard $S$ as the singleton element $s$. $\endgroup$ – David Farmilant Sep 18 '20 at 14:39

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