Show the following
[1]
Show that closure of a convex subset of a norm space is a convex set.
Solution:
Let C be convex subset of normal space N and $x,y \in \overline{C}$
Then there exists a sequence {$x_{n}$} and {$y_{n}$} in C such that
$x_{n} \longrightarrow x$ and $y_{n} \longrightarrow y$
For $ \alpha \in [0,1]$
$\alpha x_{n} + (1- \alpha)y_{n} \in C$
Since Addition and scalar multiplication is continuous so
$\alpha x_{n} + (1- \alpha)y_{n} \longrightarrow \alpha x + (1- \alpha) \in \overline{C}$
$\overline{C}$ is convex in N
$\blacksquare$
[2] Let C be a closed set in a normed linear space $\Big($$X , ||.||$ $\Big)$ over $\mathbb{F}$ , and let ($x_{n}$) be a sequence contained in C such that $\displaystyle\lim_{n \to \infty} ~ x_{n} = x \in X$.
Then $x \in C$
I'm drawing a blank and I'm not sure what I'm supposed to do here. Do I have to take an epsilon delta definition and go from there? Or somehow get a convergent sequence and show that it converges?
Can someone help here?