Closure of Convex Subset in Normed Vector Space is Convex
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Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $C \subseteq X$ be a convex subset of $X$.
Let $C^-$ be the closure of $C$.
Then $C^- \subseteq X$ is also a convex subset of $X$.
Proof
Let $x, y \in C^-$.
Suppose $x, y$ are limit points.
Then there are sequences $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that:
- $\ds \lim_{n \mathop \to \infty} x_n = x$
- $\ds \lim_{n \mathop \to \infty} x_y = y$
Let $\alpha \in \closedint 0 1$.
Then:
\(\ds \paren {1 - \alpha} x + \alpha y\) | \(=\) | \(\ds \paren {1 - \alpha} \lim_{n \mathop \to \infty} x_n + \alpha \lim_{n \mathop \to \infty} y_n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\paren {1 - \alpha} x_n + \alpha y_n}\) |
Since $C$ is convex:
- $\ds \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$
$C^-$ is closed and contains its limit points.
Hence:
- $\paren {1 - \alpha} x + \alpha y \in C^-$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces