Consider $\mathbb{R}$ with the usual metric $d$. Suppose $F \subset \mathbb{R}$, bounded from above. Show that:
i) sup $F \in \bar{F}$.
ii) If $F$ is closed in $(\mathbb{R}, d$), then sup $F \in F$.
This is what i did (using definitions from Rudin):
i) We're given that $F$ is bounded from above. This means that sup $F$ exists. Now, we show that sup $F$ is a limit point of $F$.
Let $x$ = sup $F$. Then, by definition, $\forall f \in F$, $f \le x$. If we take any neighborhood with radius $r > 0$ around $x$, since $f \le x, \forall f \in F$, there will always be an $f \in F$ s.t. $f \in N_r (x)$.
Thus, $x$ is a limit point of $F$. Since $\bar{F}$ is contains all of $F$'s limit points, it must be that $x$ = sup $F \in \bar{F}$.
ii) Is this one supposed to be simple?
$F$ is closed in ($\mathbb{R}, d$). This means $F = \bar{F}$. Since we showed (?) in i) that sup $F$ is a limit point of $F$, we know sup $F \in \bar{F}$. Since $\bar{F} = F$, we have that sup $F \in F$.
How are my approaches? I thought that, for i), showing the supremum is a limit point is the easiest way to show the supremum is inside the closure. Then, using that, we can use the definition of a closed set to show ii). But, I feel like my way of showing that sup $F$ is a limit point isn't rigorous enough. The idea was "because all elements of $F$ are all bounded by $x$, any circle around $x$ will conain some point(s) of $F$ except when the neighborhood is the singleton $x$ (but we need $r > 0$, right?). Any tips/corrections would be very helpful. Thank you.
i) You cannot conclude that x is a limit point.
For example if F is singelton
Show x is an adherance point. Whence directly, without
any misleading limit point stuff, x is in the closure.
ii) Yes it is simple. No, sup F is not always a limit point.
It is an adherance point.
x is an adherance point of A when for all open U nhood x,
U $\cap$ A is not empty. This is the correct way to define
closed sets.