MHB Proof of $A\subseteq B\implies\sup A\le\sup B$

[alexmahone]

Prove: $A\subseteq B\implies\sup A\le\sup B$

---------- Post added at 03:21 PM ---------- Previous post was at 03:03 PM ----------

By the definition of subset,

$x\in A\implies x\in B$

$\sup B$ is an upper bound of $B$.

$x\in B\implies x\le\sup B$

So, $x\in A\implies x\le\sup B$

ie $\sup B$ is an upper bound of $A$.

But $\sup A$ is the least upper bound of $A$.

So, $\sup A\le\sup B$

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Is the above proof ok?

 

[Also sprach Zarathustra]

Prove: $A\subseteq B\implies\sup A\le\sup B$

---------- Post added at 03:21 PM ---------- Previous post was at 03:03 PM ----------

By the definition of subset,

$x\in A\implies x\in B$

$\sup B$ is an upper bound of $B$.

$x\in B\implies x\le\sup B$

So, $x\in A\implies x\le\sup B$

ie $\sup B$ is an upper bound of $A$.

But $\sup A$ is the least upper bound of $A$.

So, $\sup A\le\sup B$

-----------------------------------------------------

Is the above proof ok?

Yes, but above your prove, say a word about the Axiom of completeness.

 

[issacnewton]

this looks fine to me. But the problem should state that A and B are non empty subsets of \( \mathbb{R}\) that are bounded above. Only in that case , it makes sense to talk about supremum's...

Edit: it seems somebody already said what I wanted to...hmm