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

Question

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

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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$(M_{2})$

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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!

Answer 1

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 $\inf M_2\leqslant\inf M_1$?

Answer 2

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

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

Answer 3

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...