MHB On proving sup A is less than sup B when A is in B

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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.
 
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OhMyMarkov said:
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 $\alpha \le \beta ~?$
 
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...
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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