Let me first give a definition.
By a separation of a topological space $X$, I mean a pair $U, V$ of disjoint non-empty subset of $X$ whose union is $X$.
My question revolves around this well-known theorem of connectedness in topology. Let me quote the theorem word-to-word from Munkres'.
Theorem $1$: The union of a collection of connected subspaces of $X$ that have a point in common is connected.
What is bothering me is for a set, say, $S$ to be connected, $S$ has to first be a topological space. The statement of this theorem is implicitly asserting that the union of a collection of connected subspaces of $X$ is itself a topological space.
At first I thought I misinterpreted this incorrectly. But then in the proof of Theorem $1$ above, Munkres wrote:
We (want to) prove that the space $Y = \bigcup A_\alpha$ is connected. Suppose that $Y = C \cup D$ is a separation of $Y$. The point:
To have a separation $Y$ needs to be a topological space. So, it must for sure implicitly asserting that $\bigcup A_\alpha$ is a topological space. I would like to ask whether this is a true statement until I stumbled upon this Wikipedia page which says: yes it is a topological space.
But then, reading the content of that page, I realized that that is way beyond the current scope of my topological adventure. So I would like to ask, are there gaps of my topological knowledge which I overlooked or is this a fact that needs to be taken just with faith for now?
Best Answer
There is a distinction to be made here regarding taking unions.
First, it is the definition of a topology that if $X$ is a topological space, then any arbitrary subset $U \subseteq X$ also has a topology on it called the subspace topology which is just $\mathcal{T}_U = \{A\cap U \; | \; A\subseteq X \; \text{is open}\}$. By definition, we are allowed to take an arbitrary union of sets in $X$ and this still remains a subset of $X$ imbued with the subspace topology.
However, a more advanced concept is when you take the union of distinct topological spaces. For instance, if $\{Y_\alpha \; | \; \alpha \in I\}$ is an arbitrary collection of topological spaces, then we can define the disjoint union topology on these spaces to be the union:
$$ Y \;\; =\;\; \bigcup_{\alpha \in I} Y_\alpha $$
where $S \subseteq Y$ is open if $S\cap Y_\alpha$ is open in $Y_\alpha$ for all $\alpha \in I$.