# Prove that if $M_{1} \subseteq M_{2}$, then inf$(M_{2})\leq$ inf$(M_{1})\leq$ sup$(M_{1})\leq$ sup$(M_{2})$

Prove that if $M_{1} \subseteq M_{2}$, then inf$(M_{2})\leq$ inf$(M_{1})\leq$ sup$(M_{1})\leq$ sup$(M_{2})$

My attempt:

$M_{1} \subseteq M_{2} \implies \forall m\in{M_{1}}$, $m\in{M_{2}}$

$\implies$inf$(M_{2}) \leq$ inf$(M_{1})$ and also that sup$(M_{1})\leq$ sup$(M_{2})$

Additionally, by the definition of supremum we know that inf$(M_{1})\leq$ sup$(M_{1})$

Together we have, inf$(M_{2})\leq$ inf$(M_{1})\leq$ sup$(M_{1})\leq$ sup$(M_{2})$

$\therefore$ If $M_{1} \subseteq M_{2}$, then inf$(M_{2})\leq$ inf$(M_{1})\leq$ sup$(M_{1})\leq$ sup$(M_{2})$

I made a few jumps that I am not sure you can take (line 1 to 2). Is this proof valid? Anything I can change? Thanks!

• Hmm, I think your jumps are too big. How does all m in both M_1 being in M_2 imply the inf M_2 is less or equal to the inf of M_1? How do we know from the definition of sup that inf <= sup? That's not actually part of the definition. I sympathize, as this real does seem trivial and obvious and thus irritatingly difficult to prove. But you I don't think your prove has actually done anything except restate what is to be proven and declared they are obvious. (Which to be fair they sort of are.) Jul 12 '17 at 17:40

There seems to be nothing wrong with your proof, but can you tell us how did you go from $(\forall m\in M_1):m\in M_2$ to $\inf M_2\leqslant\inf M_1$?

• This means that inf$(M_{1})$, sup$(M_{1})\in{M_{2}}$. I guess I should explicitly state that right? edit: Actually not sure that is correct Jul 12 '17 at 16:52
• @dawgchow My mistake for the earlier comment. That's not necessarily true if you think about it Jul 12 '17 at 16:52
• @dawgchow No, you shouldn't, because it is not true. Do you want a counter-example? Jul 12 '17 at 16:53
• Ya because if inf$(M_{1})$ is not the minimum of $M_{1}$ it wouldn't necessarily be in $M_{2}$ Jul 12 '17 at 16:54
• @dawgchow Right! So, your proof is incomplete. Jul 12 '17 at 16:55

Here's how I would do it:

By the definition of subsets, $\inf$ and $\sup$, we have the following:

$$M_1\subseteq M_2\iff (\forall m\in M_1)\,\,\,m\in M_2$$ $$(\forall m\in M_2)\,\,\,\inf(M_2)\le m\le \sup(M_2)$$

Then we can conclude

$$(\forall m\in M_1)\,\,\,\inf(M_2)\le m\le \sup(M_2)$$

Next, from the definition of $\inf$ and $\sup$, we get

$$(\forall m\in M_1)\,\,\,x\le m\le y \iff x\le\inf(M_1)\le\sup(M_1)\le y$$

Therefore

$$\inf(M_2)\le \inf(M_1)\le\sup(M_1)\le \sup(M_2)$$

When you are asked to prove something trivial and obvious... well, do the definitions. I think your prove relies on "common sense" and makes jumps that are simply too big, to the point you are mostly just restating the statements.

I'd do: If $m \in M_1$ then $m \in M_2$ as $M_1 \subset M_2$. So $\inf M_2 \le m$ by definition of infinum. So $\inf M_2$ is a lower bound of $M_1$. But $\inf M_1$ is the greatest lower bound of $M_1$ so $\inf M_2 \le \inf M_1$.

By identical argument: $\sup M_2 \le \sup M_1$. That is; if $m \in M_1$ then $m \in M_2$ and therefore $\sup M_2 \ge m$ and is an upper bound of $M_1$, but as $\sup M_1$ is the least upper bound $\sup M_2 \ge \sup M_1$.

By definition of supremum and infimum for any $m \in M_1$ then $\inf M_1 \le m \le \sup M_1$ so $\inf M_1 \le \sup M_1$. (Unless $M_1$ is empty, in which case neither $\inf M_1$ nor $\sup M_1$ exist.) (I suppose at the very beginning of the proof I could/should have made a comment that $M_1$ and $M_2$ are both presumed to be non-empty and bounded above and below.).

Hence ... result.

That's basically your proof but with the jumps explicitly explained. (Perhaps painfully so.)

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And, to be fair, maybe I'm being to0 glib in claiming "$\inf M_1$ is greatest lower bound".

More detail: If $\inf M_1 < \inf M_2$ then $\inf M_2$ is not a lower bound of $M_1$ which we just showed it is, so $\inf M_2 \le \inf M_1$.

That's kind of obvious but as I criticized you for not proving the obvious, it's only fair that I apply my own standards...