Notes - Metric Spaces MT23, Path-connectedness

Created: August 28, 2023

Flashcards

What does it mean for a metric space $X$ to be path-connected?
\[\forall a, b \in X \text{ } \exists \gamma : [0, 1] \to X \text{ s.t. } \gamma(0) = a, \gamma(1) = b \text{ with } \gamma \text{ continuous}\]

What is a path in a metric space $X$?

A continuous map $\gamma : [0, 1] \to X$.

Can you define the concatenation and opposite of paths $\gamma _ 1$ and $\gamma _ 2$?
\[(\gamma_1 \star \gamma_2) = \begin{cases}\gamma_1(2t) &0 \le t \le \frac 1 2 \\\\ \gamma_2(2t - 1) &\frac 1 2 \le t \le 1\end{cases}\] \[\gamma^{-}(t) = \gamma(1 - t)\]

What are the equivalence classes of the equivalence relation $a \sim b$ for $a, b \in X$ where $X$ is a metric space and $\sim$ is “there exists a path between them”?

The path-components of $X$.

What implication links connectedness and path-connectedness?
\[\text{path-connectedness} \implies \text{connectedness}\]

In general, connectedness does not imply path-connectedness. But what theorem lets this happen sometimes?

Suppose:

  • $X$ is a normed space
  • $U$ is a connected, open subset of $X$

Then:

\[U \text{ connected} \iff U \text{ path-connected}\]

Can you give an example of where connectedness does not imply path-connectedness?
\[A = \\{(0, y) : -1 \le y \le 1\\} \cup \\{(x, \sin(1/x)) : x \in (0, 1]\\}\]

What is true about the connectedness of any normed vector space (over $\mathbb R$ or $\mathbb C$)?

It is path-connected.

Quickly prove that if

  • $U \subseteq \mathbb R^n$, open and connected

then

  • $U$ is path-connected

Fix some $a \in U$ and let $\Gamma(a)$ denote the path-component containing $a$. We aim to show that

\[K := U \setminus \Gamma(a)\]

is empty. Note $\Gamma(a)$ is open, since for any $x \in \Gamma(a)$, $\exists \varepsilon > 0$ such that $B(x, \varepsilon) \subseteq U$. Then for any $y \in B(x, \varepsilon)$ there is a straight-line path in $\mathbb R^n$ that connects $x$ and $y$, which implies that $a$ and $y$ are path-connected (transitivity). So $y \in \Gamma(a)$, hence $B(x, \varepsilon) \subseteq \Gamma(a)$.

Similarly $K$ is open. For any $x \in K$, $\exists \varepsilon > 0$ such that $B(x, \varepsilon) \subseteq U$. But then if there were any $y \in B(x, \varepsilon)$ that connected $y$ and $a$, $x$ and $a$ would be connected, which contradicts the fact $x \in K$. So $B(x, \varepsilon) \subseteq K$.

By the definition of set difference, $K \cup \Gamma(a) = U$ and $K \cap \Gamma(a) = \emptyset$. Since $U$ is non-empty, if $K$ was non-empty, $K$ and $\Gamma(a)$ would disconnect $U$, a contradiction. So $K$ is empty.

Suppose:

  • $X$ is path-connected

Quickly prove that:

  • $X$ is connected

We use the following characterisation of connectedness:

$X$ is connected $\iff$ every continuous function $f : X \to \{0, 1\}$ is constant.

Suppose $f : X \to \{0, 1\}$ is a continuous function. Pick two arbitrary points $x, y \in X$, we aim to show $f(x) = f(y)$. By assumption, there exists a continuous map $\gamma : [0, 1] \to X$ such that $\gamma(0) = x, \gamma(1) = y$.

Then the composition $f \circ \gamma : [0, 1] \to \{0, 1\}$ is continuous, and therefore constant (since $[0, 1]$ is connected). Hence

\[f(x) = (f \circ \gamma)(0) = (f \circ \gamma)(1) = f(y)\]

So $f$ is a constant function.

Alternative proof: Suppose $X$ is not connected. Then $\exists U, V$ non-empty disjoint open subsets $U, V \subseteq X$ such that $X = U \cup V$. Pick some $x \in U$ and $y \in V$.

Then $\exists f : [0, 1] \to X$ such that $f(0) = x$ and $f(1) = y$.

Consider $f^{-1}(U)$ and $f^{-1}(V)$. These are disjoint in $[0, 1]$ and their union is $[0, 1]$ (since $X = U \cup V$).

By the continuity of $f$, they are both open in $[0, 1]$.

What’s a useful trick when trying to deduce if a subset of a metric space is path-connected?

Rather than considering arbitrary points $x$ and $y$, you can just consider if it’s possible to map some arbitrary $x$ to a reference point (like $\pmb 0$), since then concatenating opposite paths gives a general connection between any two points.



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