1
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ Do you know an example of a connected space with two path components? $\endgroup$ – Eric Wofsey Oct 10 '17 at 6:02
  • $\begingroup$ 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? $\endgroup$ – PrimeSpiralObserver Oct 10 '17 at 6:11
  • 2
    $\begingroup$ path components do not have to be closed (finite or not). $\endgroup$ – CopyPasteIt Oct 14 '17 at 0:34
  • $\begingroup$ $(-\infty,1]\cup [2,3]\cup [5,\infty)$ with subspace topology induced from $\mathbb{R}$. $\endgroup$ – user 170039 Jul 10 at 6:39
6
$\begingroup$

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 $23.4$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.