Nonempty topologically convex implies equiconnected

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., nonempty topologically convex space) must also satisfy the second topological space property (i.e., equiconnected space)
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Statement

Topological version

Any nonempty topologically convex space is an equiconnected space.

Realized version

Any nonempty convex subset of Euclidean space is (topologically) an equiconnected space.

Definitions used

Nonempty topologically convex space

Further information: nonempty topologically convex space

A topological space is called a nonempty topologically convex space if it is nonempty and homeomorphic to a convex subset of Euclidean space.

Convex subset of Euclidean space

Further information: convex subset of Euclidean space

A convex subset of Euclidean space is a subset in $\mathbb {R} ^{n}$ for some $n$, with the property that given any two points in the subset, the line segment joining those two points also lies completely within the subset.

The definition may also apply to infinite-dimensional Euclidean spaces, and the proof would work even in that case.

Equiconnected space

Further information: equiconnected space

A nonempty topological space $X$ is said to be equiconnected if there is a continuous map $k:X\times [0,1]\times X\to X$ such that $k(x,t,x)=x$ for all $x$ and $k(x,0,y)=x,k(x,1,y)=y$ for all $x,y\in X$.

Proof

Given: A convex subset $X$ of $\mathbb {R} ^{n}$

To prove: There is a continuous map $k:X\times [0,1]\times X\to X$ such that $k(x,t,x)=x$ for all $x$ and $k(x,0,y)=x,k(x,1,y)=y$ for all $x,y\in X$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation	Geometric interpretation
1	Define $k:X\times [0,1]\times X\to X$ as follows: $k(x,t,y):=(1-t)x+ty$. Here, the scalar multiplication addition is as vectors in $\mathbb {R} ^{n}$.		$X$ is a convex subset of $\mathbb {R} ^{n}$		$k(x,t,y)$ is a convex combination of $x$ and $y$, so by convexity of $X$, it is inside $X$.	$k$ defines the point on the line segment joining the points $x$ and $y$ that divides the line segment in the ratio $t:(1-t)$. By convexity of $X$, this point is inside $X$. As $t$ increases from 0 to 1, $k(x,t,y)$ moves at constant speed along the line segment from $x$ to $y$.
2	$k$ is continuous.			Step (1)	This follows from addition and scalar multiplication being continuous functions, and the fact that the composite of continuous functions is continuous.	This is tricky to visualize (even for something simple such as a unit interval), so we shall skip the description.
3	$k(x,t,x)=x$ for all $x\in X,t\in [0,1]$.			Step (1)	This follows by plugging in $y=x$ in the definition in Step (1).	When $x=y$, the "line segment" joining $x$ and $y$ becomes a point, so for all $t$ we stay at that point.
4	$k(x,0,y)=x$ for all $x,y\in X$.			Step (1)	This follows pretty directly by plugging $t=0$ in the expression for $k$ in Step (1).	This is the "start" (at $t=0$) of the traversal along the line segment from $x$ to $y$, so it is naturally at $x$.
5	$k(x,1,y)=y$ for all $x,y\in X$.		Step (1)	This follows pretty directly by plugging $t=1$ in the expression for $k$ in Step (1).	This is the "end" (at $t=1$) of the traversal along the line segment from $x$ to $y$, so it is naturally at $y$.
6	$k$ is the desired function.			Steps (1), (2), (3), (4), (5)	Step (1) defines $k$ with the correct domain and co-domain. Steps (2), (3), (4), and (5) establish the criteria that $k$ needs to satisfy.

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