Prove convex hull is a compact set

convex-analysisconvex-hullslinear algebra

Let $\mathbf{v}_1,\dots,\mathbf{v}_r$ be vectors in a Euclidean space $\mathbf{V}$.

Prove that the convex hull
$\mathrm{Conv}(\mathbf{v}_1,\dots,\mathbf{v}_r)$ is a compact set.

I believe the convex hull is defined as the span of the vectors in V whose scalar coefficients are nonnegative and sum to 1. To prove it's compact, one must prove that the set is both closed and bounded but I'm not sure how to do that.

Best Answer

If we define $C = \{(x_1, \dots, x_r) \in \Bbb R^r: x_i \geq 0, \sum x_i = 1\}$, then the convex hull of $v_1, \dots, v_r \in V$ is nothing but the image of the map $f: C \rightarrow V$ sending $(x_1, \dots, x_r)$ to $\sum x_i v_i$.

The map $f$, being a linear map (if we extend the domain to $\Bbb R^r$), is clearly continuous. Thus it suffices to show that $C$ is compact in $\Bbb R^r$, as the image of a compact set under a continuous map is again compact.

So it suffices to show that

  1. $C$ is closed;
  2. $C$ is bounded.

Both are pretty obvious.

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