Let Y be non empty subset of R which is bounded above and y = supY then show that y $\in Y \cup$ Y'.

Let Y be non empty subset of R which is bounded above and y $=$ supY then show that y $\in Y \cup$ Y', where Y' is set of limit points of Y

Now sup exists by lub property. If sup belongs to set Y then i am done. if not then there exists an element a $\in$ Y such that $a > y- \epsilon$. so a $\in$ $(y-\epsilon, y + \epsilon$ ) for any spsilon. So neighbourhood of y has a element a. So it is limit point.

Is this correct?

Thanks

• If $Y$ is a non empty subset of $\Bbb R$ then I assume that the order relation is $\le$ and not $\subseteq$, so how is possible that $y\in \sup(Y)$ if $\sup(Y)$ is a number??? – Masacroso Sep 1 '16 at 14:57
• As Masacroso pointed out, $\sup Y$ is not a set, so $y\in\sup Y$ makes no sense. The correct theorem is that if $y\color{red}=\sup Y$, then $y\in Y\cup Y'$; your argument is a correct proof of this statement. – Brian M. Scott Sep 1 '16 at 15:20
• Well, technically, every real number is a set... But of course the remarks in the pair of comments above fully stand. – Did Sep 1 '16 at 15:22
• @BrianM.Scott How do i write this in terms of metric space and neighbourhoods – J. Deff Sep 1 '16 at 15:26
• @J.Deff: I’m not sure what you mean: $(y-\epsilon,y+\epsilon)$ is the $\epsilon$-nbhd of $y$ when $\Bbb R$ is given the usual metric. – Brian M. Scott Sep 1 '16 at 15:29