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[SOLVED] On proving sup A is less than sup B when A is in B

OhMyMarkov

Member
Mar 5, 2012
83
Hello everybody!

I want to prove that if $A\subset B$, then $\sup A \leq \sup B$. I'm taking the exhausting approach of considering cases in proving this:

First let $\alpha = \sup A, \beta = \sup B$

(1) If $\alpha \in A, \alpha \in B,$ so $\alpha \leq \beta$

(2) If $\alpha \notin A, \alpha \in B$ the $\sup$ of $B$ is bigger than all elements in $B$, nameley $\alpha$, so $\alpha \leq \beta$

(3) If $\alpha \notin A, \alpha \notin B$, now there seems to be two subcases here:
a- if $\alpha < \beta$
b- if $\alpha = \beta$

But I can't seem to establish those!
Any help on that is appreciated, if there are shortcuts or a quicker proof I'd be thankful if I can see it.
 
Last edited:

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
I want to prove that if $A\subset B$, then $\sup A \leq \sup B$. I'm taking the exhausting approach of considering cases in proving this:
First let $\alpha = \sup A, \beta = \sup B$

First let us assume that $\beta = \sup B$ actually exists, i.e. $B$ has an upper bound.
By the given $\alpha = \sup A$ must then also exist.

Suppose that $\beta < \alpha$. That means that $\beta$ is not an upper bound of $A$ WHY?

How is that a contradiction?

How does that prove that $\alpha \le \beta ~?$
 

OhMyMarkov

Member
Mar 5, 2012
83
Ok...

(1) Suppose $\beta < \alpha$, since $\alpha = \sup A$, any number $t < \alpha$ is not an upper bound of $A$ by the definition of the least upper bound

(2) But $\beta$ is an upper bound of $B$, so $\forall x \in B$, $x \leq \beta$, in particular, every $x\in A, x \leq \beta$ so that $\beta$ is also an upper bound for $A$!

A contradiction!

(3) Hence, $\alpha \leq \beta$

Thank you, I think I got it right...