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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?

saru
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    Munkres is referring to the subspace topology. –  Apr 16 '20 at 21:09
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    Indeed: all of the subsets of $X$, including the union of the collection of subspaces, are to be endowed with the subspace topology. The background fact that makes this a consistent interpretation is that if $V\subset W\subset X$ are topological spaces, then the subspace-of-$X$ topology on $V$ is the same topology as the subspace-of-$W$ topology on $X$, when $W$ itself inherits the subspace-of-$X$ topology. – Greg Martin Apr 16 '20 at 21:10
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    I think that's the right way to look at it. To expand: the union will be some subset of the original topological space, and, as such, can be given the subspace topology. Munkres's claim is that the union is connected in the subspace topology. – Charles Hudgins Apr 16 '20 at 21:11
  • @CharlesHudgins your comment gave me the most insight to answering my confusion. I would definitely accept your answer if it's not a comment. – saru Apr 16 '20 at 21:53
  • @GregMartin I found your comment very much useful as well. Thank you. – saru Apr 16 '20 at 21:54

3 Answers3

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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 $\alpha \in I$.

Mnifldz
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    Thank you for pointing out the distinction between this more advanced concept of disjoint union topology and the subspace topology. – saru Apr 16 '20 at 21:52
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    I'd add that the disjoint union of (two or more non-empty) topological spaces is always disconnected. (The theorem in the question is still true if having a point in common means a quotient where we identify one point from each space, but that's probably too much of a digression.) – Joonas Ilmavirta Apr 17 '20 at 08:56
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"...collection of connected subspaces of $X$..." means that these have the topology inherited from $X$, i.e., the subspace topology, as does their union.

MPW
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Munkres is referring to the subspace topology, i.e. given a topological space $(X,\tau)$ and a subset $S\subseteq X$, the subspace topology on $S$ is $\tau_S=\{A\cap S\,:\, A\in\tau\}$.