# Find a topological space $X$ which is connected but has three path components.

What would a connected topological space $X$ look like with three path components? I know that since it has a finite number of path components, these components are closed but I'm not sure if that helps.

• Do you know an example of a connected space with two path components? – Eric Wofsey Oct 10 '17 at 6:02
• If a point $x ∈ X$ and $P(x)$ is the subspace of $X$ consisting of points $y$ such that there is a path in $X$ from $x$ to $y$, would a point $z ∈ P(x) ∩ P(y)$, have two path components? – PrimeSpiralObserver Oct 10 '17 at 6:11
• path components do not have to be closed (finite or not). – CopyPasteIt Oct 14 '17 at 0:34
• $(-\infty,1]\cup [2,3]\cup [5,\infty)$ with subspace topology induced from $\mathbb{R}$. – user 170039 Jul 10 at 6:39

One can find examples as subspaces of $$\mathbb{R}^n$$. Try taking advantage of the fact that any set between (in the sense of containment) a connected set and its closure—up to the closure itself—is also connected*. It can be the case, when $$n>1$$, that the closure of a path-connected (hence connected) set $$X$$ is no longer path-connected, which means we can potentially increase the total number of path components by passing from a connected set to its closure. Subsequently, there's a chance we can get even more by deleting points from path components in $$\overline{X} \setminus X$$, all the while preserving connectedness.

For instance, consider the path-connected set $$X = \left\{ \big(x, \sin(1/x) \big) \ | \ x \in (0, 1] \right\} \subset \mathbb{R}^2$$ (cf. the topologist's sine curve). What is $$\overline{X}$$? You'll find that $$\overline{X}$$ is no longer path-connected, having two path components. Moreover, by deleting points from $$\overline{X} \setminus X$$, we can construct a set $$Y$$, where $$X \subset Y \subset \overline{X}$$, such that $$Y$$ has arbitrarily many path components (even as many as $$\aleph_0$$ or $$\aleph_1$$) while still being connected.

*For proof, see Munkres' Topology: Chapter $$3$$, theorem $$23.4$$

The following is easy to prove:

Proposition 1: Let $X$ be any topological space with a nested decreasing chain of nonempty open subsets:

$\tag 1 U_0 \supset U_1 \supset \dots \supset U_n \supset \dots$

Let $\rho$ be any element not in $X$. Then the collection of sets $\mathcal B$ defined by

$\tag 2 U \in \mathcal B \text{ iff [} U \text{ is open in } X \text{ or } U = U_n \cup \{\rho\} \text{]}$

forms a topological basis for the set $\hat X = X \cup \{\rho\}$.

For the topological space $\mathbb R$ we can define the decreasing chain of open sets

$\tag 3 U_n = \mathbb R - \mathbb Z - \{x \in \mathbb R \, | \; |x| \lt n \}$

and can therefore create the space $\hat {\mathbb R}$ as shown above, which is also a Hausdorff space.

Proposition 2: The space $\hat {\mathbb R}$ is connected with two path components.
Proof
It is easy to argue that any clopen containing $\rho$ is all of $\hat{\mathbb X}$, so we have a connected space.

We show that if $\gamma$ is any path with $\gamma(0) \in \mathbb R$, then the image $\gamma([0,1])$ is also contained in $\mathbb R$. Recall that the image any path is connected.

If not, there is a map $\gamma^{'}$ with $\gamma^{'}([0,1)) \subset (n, n+1)$ and $\gamma^{'}(1) = \rho$. But for continuous functions the image of the closure is contained in closure of the image, yet $\rho \notin [n, n+1]$. $\qquad \blacksquare$

Let there be given two disjoint copies $\mathbb R_1$ and $\mathbb R_2$ of the number line and take $\mathbb X$ to be the disjoint sum of these spaces. We will now define a decreasing chain ${\mathbb U}_n$ of open sets in $\mathbb X$. For $i \in \{1,2\}$, set

$\tag 4 {U_n}^i = \mathbb R_i - \mathbb Z_i - \{x \in \mathbb R_i \, | \; |x| \lt n \}$

and

$\tag 5 {\mathbb U}_n = {U_n}^1 \sqcup {U_n}^2$

So by proposition 1 we have a space $\hat{\mathbb X}$.

Proposition 3: The space $\hat{\mathbb X}$ is connected with three path components.
Proof
Let a path $\gamma: [0,1] \to \hat{\mathbb X}$ be given, and suppose that, say, $\gamma(0) \in \mathbb R_1$. By continuity, there is an $a \gt 0$ such that the image of the interval $[0,a)$ under $\gamma$ is contained in $\mathbb R_1$. If $\gamma$ maps any points outside of $\mathbb R_1$ then there is a map $\gamma^{'}$ with $\gamma^{'}([0,1))$ contained in $\mathbb R_1$ and $\gamma^{'}(1) \in \mathbb R_2 \cup \{\rho\}$. But then in either case we can get a contradiction. $\qquad \blacksquare$

This argument can be extended to show that we can construct a connected space with an arbitrary number of path components (Kaj Hansen also points this out). In this construction exactly one path component is a singleton set.