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#### 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.

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.

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