> [!definition] > > Let $\cx$ be a [[Vector Space|vector space]]. A subset $C \subset X$ is **convex** if > $ > \forall x, y \in C, \bracs{tx + (1 - t)y: t \in [0, 1]} \subset C > $ > [!theorem] > > Let $\cx$ be a [[Vector Space|vector space]], $x, y \in \cx$ and $a, b > 0$, then > $ > \exists t_0 \in (0, 1): \frac{x + y}{a + b} = \frac{t_0x}{a} + \frac{(1 - t_0)y}{b} > $ > *Proof*. Let $t_0 = \frac{a}{a + b}$, then > $ > \frac{t_0x}{a} + \frac{(1 - t_0)y}{b} = \frac{x}{a + b} + \frac{y}{a + b} = \frac{x + y}{a + b} > $ > [!theorem] > > Let $X$ be a vector space and $E \subset X$, then $\text{conv}(E)$ is the smallest convex set containing $E$, and > $ > \text{conv}(E) = \underbrace{\bracs{\sum_{x \in E'}a_xx: E' \subset E, E' \text{ finite}, \sum_{x \in E'}a_x = 1}}_{\ce} > $ > *Proof*. Firstly, $\ce$ itself is convex, as a convex combination of two convex combinations is still a convex combination: > $ > t\sum_{x \in E'}a_xx + (1 - t)\sum_{x \in E''}b_xx > $ > is a linear combination of elements of $E$, with > $ > t\sum_{x \in E'}a_x + (1 - t)\sum_{x \in E''}b_x = t + (1 - t) = 1 > $ > Now, we show that $\text{conv}(E)$ contains all convex combinations of $E$ by induction on the size of the subset. For $|E'| = 2$, > $ > tx + (1 - t)y \in E \quad \forall \bracs{x, y} = E' \subset E > $ > Suppose that any convex combination of $n$ vectors is an element of $E$, and let $\sum_{k = 1}^{n}a_kx_k + by$ be a convex combination of $n + 1$ vectors, then > $ > \begin{align*} > \sum_{k = 1}^{n}a_kx_k + by &= ty + (1 - t)\underbrace{\sum_{k = 1}^{n}\frac{a_k}{1 - t}x_k}_{\in E} \in E > \end{align*} > $ > Therefore $\text{conv}(E) = E$. > [!theorem] > > Let $\cx$ be a normed vector space and $C$ be a convex set, then $\ol{C}$ is also convex. > > *Proof*. Let $\varepsilon > 0$, $x, y \in \ol{C}$, and $x', y' \in C$ such that $\norm{x' - x}, \norm{y' - y} < \varepsilon$. Then > $ > \begin{align*} > &\norm{(tx' + (1 - t)y') - (tx + (1 - t)y)} \\ > &\le t\norm{x' - x} + (1 - t)\norm{y' - y} \\ > &< t\varepsilon + (1 - t)\varepsilon = \varepsilon > \end{align*} > $ > for all $t \in [0, 1]$. Since this applies to all $\varepsilon > 0$, $tx + (1 - t)y \in \ol{C}$ for all $t \in [0, 1]$.