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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

---------- 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?