Convex Set is Contractible

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Theorem

Let $V$ be a topological vector space over $\R$ or $\C$.

Let $A\subset V$ be a convex subset.

Then $A$ is contractible.

Proof

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Let $x_0 \in A$.

Define $H : A \times \closedint 0 1 \to A$ by:

$\map H {x, t} = t x_0 + \paren {1 - t} x$

Note that by the definition of a convex set, $\Cdm H$ is also a convex set.

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Let $I_A$ be defined as the identity mapping:

$I_A: A \to A$

Such that:

$I_A = \set {\tuple {x, x}: x \in A}$

Let $S$ and $T$ be sets. Let $x_0$ be defined as the constant map:

$x_0: S \to T$ defined as:

$0 \in T: x_0: S \to T: \forall x \in S: \map {x_0} x = 0$

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The map $H$ yields a (free) homotopy between the identity map:

$I_A$

and the constant map:

$x_0$

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By the assumption of convexity for $A$, $H$ takes values in $A$.

Since $H$ is a polynomial separately in $x, t$, it is a continuous function.

And:

$\map H {-, 0} = I_A$

$\map H {-, 1} \equiv x_0$ (the constant function)

Thus, $H: I_A \simeq c_{x_0}$.

This needs considerable tedious hard slog to complete it. In particular: Prove that the above statement implies that the identity map on A is homotopic to the constant map from A to A To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

It follows that, the identity map $\operatorname{id}_A$ is homotopic to the constant map $A \to A$.

Hence, by definition of a contractible space, this implies that:

$A$ is contractible

as required.

$\blacksquare$