How can I prove that the union of a collection of continuous functions is a topological space?

Question

Let $I = [0,1]$ be given the topology induced by the standard topolgy of $\mathbb{R}$. Let $X$ be a topological space. We say that $X$ is path-connected if, for all $x,y \in X$ there exists a continuous function: $\gamma:I \to X$ such that $\gamma(0) = x$ and $\gamma(1) = y$, we call $x$ and $y$ the start and end points of the path: $\gamma$ respectively.

I want to prove that there exists a collection: $\Gamma$ of paths in $X$, all sharing the same start point, such that: $X = \bigcup_{\gamma \in \Gamma} \gamma(I)$. However I am unsure how to prove that the union of continuous functions is a topological space.

Answer 1

Assuming that $X\neq\emptyset$, fix some $x\in X$. For each $y\in X$, let $\gamma_y$ be a path starting in $x$ and ending in $y$. Then$$X=\bigcup_{y\in X}\gamma_y(I),$$since, for each $y\in X$, $y=\gamma_y(1)\in\bigcup_{y\in X}\gamma_y(I)$.

Answer 2

A "union of continuous functions" can be made into a topological space in many ways. But this is not the problem before your eyes. The problem is to prove that two explicitly described sets are equal. In the case at hand this is obvious.