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