Other approaches to connected components (from this year's MATH4085 Metric and Topological Spaces)

Hi everyone,

In a previous post we looked at a sufficient condition for a union of connected sets to be connected. A relatively easy special case of Lemma 6.12 is the following.

Lemma
Let \(X\) be a (non-empty) topological space, and let \(E_1\) and \(E_2\) be connected subsets of \(X\) such that \(E_1 \cap E_2 \neq \emptyset\,.\) Then \(E_1 \cup E_2\) is also connected.

Armed just with this lemma, we can define an equivalence relation  \(\sim\) on a non-empty topological space \((X,\tau)\) as follows: for \(x,y \in X\), \(x\sim y\) if and only if there exists a connected subset \(E\) of \(X\) such that both \(x\) and \(y\) are elements of \(E\).

Let's check that this really is an equivalence relation.

Since single point sets \(\{x\}\) are connected, reflexivity is easy. Symmetry is also immediate (it doesn't matter in which order you say that both \(x\) and \(y\) are elements of \(E\)).

That leaves transitivity, which is where we use the lemma above.

Let \(x,y,z \in X\) and suppose that \(x \sim y\) and that \(y \sim z\). Since \(x \sim y\), there is a connected subset \(E_1\) of \(X\) such that \(x\) and \(y\) are elements of \(E_1\). Similarly, since \(y \sim z\), there is a connected subset \(E_2\) of \(X\) such that \(y\) and \(z\) are elements of \(E_2\). Set \(E=E_1 \cup E_2\). Note that \(y \in E_1 \cap E_2\), so this intersection is non-empty. By the lemma above, \(E\) is connected. But \(x\) and \(z\) are both in \(E\), so \(x \sim z\).

What are the equivalence classes? They are just the connected components of \(X\) again. If we assume Lemma 6.12 and use that to define connected components, this is obvious: we see that \(x \sim y\) if and only if \(x\) and \(y\) are in the same connected component of \(X\).

Otherwise, if we just work with the weaker lemma above, we should prove that each equivalence class is connected. So, let \(x_0 \in X\), and set

\(K=[x_0]=\{y \in X: x_0\sim y\}\,.\)

As usual, we give \(K\) the subspace topology \(\tau_K\). We show that \((K,\tau_K)\) is connected. 

We first claim the following useful fact (that is valid outside this proof too, with \(K\) and \(x_0\) as above): whenever \(E\) is a \(\tau\)-connected subset of \(X\) with \(x_0 \in E\), then \(E \subseteq K\). To see this, note that for each \(y \in E\) we have both \(x_0\) and \(y\) are in \(E\), and so \(x_0 \sim y\), and hence \(y \in [x_0] = K\), as required.

Now let \(A\) be a \(\tau_k\)-clopen subset of \(K\).

First suppose that \(x_0 \in A\). We show that in this case \(A=K\).

Let \(y\in K\). Then there is a \(\tau\)-connected subset \(E\) of \(X\) such that \(x_0\) and \(y\) are both in \(E\). By our claim above we have that \(E \subseteq K\). But then \(A \cap E\) is a \(\tau_E\)-clopen subset of \(E\), and is non-empty (\(x_0\) is an element). Thus we must have \(E \subseteq  A\) (as usual, via \(A \cap E = E\)). Finally, this shows that \(y \in A\).

This is true for all \(y \in K\),  so we have \(K \subseteq A\), and equality holds.

The remaining case is when \(x_0 \notin A\). But then \(K\setminus A\) is also \(\tau_K\)-clopen in \(K\), and \(x_0\in K\setminus A\), so the above argument shows that \(K\setminus A=K\), and hence \(A=\emptyset\).

Thus the only \(\tau_K\)-clopen subsets of \(K\) are \(K\) and \(\emptyset\). This shows that \(K\) is connected.

We have that \(K\) is a connected subset of \(X\) which contains \(x_0\). Also, as we showed above, whenever \(E\) is a connected subset of \(X\) such that \(x_0 \in E\), we have \(E \subseteq K\). So \(K\) really is the maximum possible connected subset of \(X\) which has \(x_0\) as an element.

Either approach works just as well! But perhaps the full version of Lemma 6.12 saves some work.

Best wishes,
Dr Feinstein