Connectedness of unions (from this year's MATH4085 Metric and Topological Spaces)

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In order to formalise the notion of "connected components", we need to know something about when a union of connected sets might be connected. Of course this isn't generally true, e.g. (with the usual topology)

\((0,1) \cup (2,3)\)

is disconnected, but is a union of two connected sets. On the other the union of two disjoint connected sets can sometimes be connected. For example,

\((0,2)=(0,1]\cup(1,2)\)

is connected, even though it is a union of two pairwise disjoint connected sets.
[Can you work out what the true story is here?]

Let's have a look at Lemma 6.12 which gives a sufficient condition for a union of connected sets to be connected. 


Informally, these connected pieces are somehow all "attached" to each other at the point \(x_0.\) The proof of Lemma 6.12 makes this formal.

There are many ways to prove this lemma. For example, we can assume that the union is equal to \(X\) if we like. We can also use a variety of conditions equivalent to connectedness. But here we will work with the union directly and with the definition using clopen sets.

We may assume that \(I\neq \emptyset\), because the empty union is the empty set, which is connected. 

Set

\(\displaystyle X'=\bigcup_{i\in I} E_i\,,\)

and give \(X'\) the subspace topology \(\tau_{X'}\) induced on \(X'\) by the topology on \(X\). Note that \(x_0 \in X'\) (as long as we assume that \(I \neq \emptyset\)).

For each \(i \in I\) we give the set \(E_i\) its subspace topology (from \(X\) or from \(X'\), it doesn't matter), which we denote by \(\tau_i\).
[You might remember that we said in the notes that the subsubspace topology is the same as the subspace topology?]

We know that each \((E_i,\tau_i)\) is a connected topological space.
[For the rest of the proof we work either in the topological space \((X',\tau_{X'})\) or in the topological spaces \(E_i\) with their subspace topologies.]

We have to show that \((X',\tau_{X'})\) is connected.

Let \(A\) be a \(\tau_{X'}\)-clopen subset of \(X'\).
[We show that \(A\) must be either \(\emptyset\) or \(X'\). To detect which, we consider separately the two cases where \(x_0 \in A\) and where \(x_0 \notin A\).]

Case I: Suppose that \(x_0\in A\).

Then, for each \(i \in I\), we have \(x_0 \in A \cap E_i\) and so, in particular, \(A \cap E_i \neq \emptyset\).

But now \((E_i,\tau_i)\) is connected and \(A \cap E_i\) is a non-empty \(\tau_i\)-clopen subset of \(E_i\). By connectedness, we must have \(A \cap E_i=E_i\), i.e., \(E_i \subseteq A.\)

This is true for all \(i\in I\). Thus we have

\(\displaystyle X'=\bigcup_{i\in I} E_i \subseteq A \subseteq X',\)

and so \(A=X'.\)

Case II: Otherwise we have \(x_0 \in X' \setminus A\).

But \(X'\setminus A\) is also a \(\tau_{X'}\)-clopen subset of \(X'.\) So the argument above shows that \(X'\setminus A=X'\), and so \(A=\emptyset\).

Thus the only \(\tau_{X'}\)-clopen subsets of \(X'\) are \(\emptyset\) and \(X'\), as required. \(\quad\square\) 

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