# Prove that $\inf S \le \inf A \le \sup A \le \sup S$.

Let $$S \subset \mathbb{R}$$ (real numbers) be a bounded subset and let $$A \subset \mathbb{R}$$ be a non-empty subset of $$S$$. Prove that $$\inf S \le \inf A \le \sup A \le \sup S$$.

Having trouble understanding this. I get that obviously the $$\inf S$$ and $$\inf A$$ will be less than $$\sup A$$ and $$\sup S$$, but I don't understand why $$S$$ and $$A$$ are different. If I could visualize $$S$$ and $$A$$ on a number line that would help a lot. And then to prove this I can't right now as I need more understanding.

• Some simple examples to guide your thinking: $S = [1,2] \subset [0,3] = A$, $S = (0,1) \subset [0,1] = A$ and $S = (0,1) \cup (2,3) \subset [0,9] = A$. – BaronVT Sep 10 '19 at 0:50
• The thing is if $k$ is a lowerbound of $S$ it must be a lower bound of $A$ because $A$ is subset of $A$ and if $k$ is less than every element of $S$ then it is lower than every element of $A$ because every element of $A$ is an element of $S$.... The reverse isn't true because if $m$ is a lower bound of $A$ and $m$ is lower than every element of $A$, there could be elements in $S$ that are not in $A$ that are much lower.... An example could but $A = (2,3) \subset (1,4)= S$. So $\inf S=1 < \inf A=2< \sup A =3 < \sup S = 4$. – fleablood Sep 10 '19 at 1:11
• I am confused, how is $A = (2,3) \subset (1,4) if neither of the elements in A are in S? – Richard Smith Sep 10 '19 at 1:16 •$A$is the open interval$(2,3)$. It is all the numbers between$2$and$3$.$S$is the open interval$(1,4)$. It is all the numbers between$1$and$4$. If$x \in A$then$2 < x < 3$and so$1 < 2 < x < 3 < 4$so$1 < x < 4$. So$x \in S$. Thus$A\subset S\$. – fleablood Sep 10 '19 at 1:26

The thing is if $$k$$ is a lowerbound of $$S$$ it must be a lower bound of $$A$$ because $$A$$ is subset of $$A$$ and if $$k$$ is less than every element of $$S$$ then it is lower than every element of $$A$$ because every element of $$A$$ is an element of $$S$$.... The reverse isn't true because if $$m$$ is a lower bound of $$A$$ and $$m$$ is lower than every element of $$A$$, there could be elements in $$S$$ that are not in $$A$$ that are much lower.... An example could but $$A = (2,3) \subset (1,4)= S$$. So $$\inf S=1 < \inf A=2< \sup A =3 < \sup S = 4$$.

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To do this formally just use the definitions.

Given: $$S$$ is bounded above and below. As $$\mathbb R$$ has the least upper and lower bound property. $$\inf S$$ and $$\sup S$$ exist.

Claim: $$A$$ is bounded above and below.

Pf: $$S$$ is bounded below so there is a $$k$$ so that $$k \le s$$ for every $$s\in S$$. For any $$a\in A$$, $$a \in S$$ because that's what a subset means. So $$k\le a$$ becasue $$k$$ is less or equal to every element of $$S$$. So $$A$$ is bounded below.

So $$A$$ is bounded below. THe same argument holds to show $$A$$ is bounded above. And by l.u.b. property $$\inf A$$ and $$\sup A$$ exist.

Claim: $$\inf S \le \inf A$$.

I'll leave the proof to you but it is a similar subset argument.

$$\inf S$$ is a lower bound of $$S$$. Show that means $$\inf S$$ is a lower bound of $$A$$. For any $$k; k > \inf A$$ then $$k$$ can not be a lower bound of $$A$$ so show it can't be a lower bound of $$S$$ either. What happens if $$\inf S > \inf A$$?

Claim