# Notes - Metric Spaces MT23, Connectedness

### Flashcards

What does it mean for a metric space $X$ to be disconnected?

It can be written as the disjoint union of two nonempty open sets.

What has to be true about the two sets whose union form $X$ if it is disconnected?

- Nonempty
- Open
- Disjoint

What are three equivalent characterisations of connectedness for a metric space $X$?

- $X$ is connected
- If $f : X \to \{0, 1\}$ is continuous, it is constant.
- The only open and closed subsets of $X$ are $X$ and $\varnothing$.

If $f : X \to \{0, 1\}$ is a continuous function on a disconnected metric space $X = A \cup B$, where $A \cap B = \emptyset$ and $A, B$ open, what sort of function is $f$?

or

\[f(x) = \mathbb 1_B = 1-\mathbb 1_A\]Can you state the lemma that lets you characterise whether a subset $Y \subseteq X$ is connected if and only if some condition involving open subsets $U, V$ of $X$ holds?

Suppose:

- $Y \subseteq X$
- $U, V \subseteq X$, open
- $U \cap V \cap Y = \emptyset$

Then $Y$ is connected if the following condition holds:

- $Y \subseteq U \cup V$

iff

- $Y \subseteq U$ or $Y \subseteq V$

What does the following lemma “really say”:

Let $X$ be a metric space and $Y \subseteq X$. Then $Y$ is connected if and only if the following is true: If $U, V$ are open subsets of $X$ and $U \cap V \cap Y = \varnothing$, then whenever $Y \subseteq U \cup V$, either $Y \subseteq U$ or $Y \subseteq V$.

Let $X$ be a metric space and $Y \subseteq X$. Then $Y$ is connected if and only if the following is true: If $U, V$ are open subsets of $X$ and $U \cap V \cap Y = \varnothing$, then whenever $Y \subseteq U \cup V$, either $Y \subseteq U$ or $Y \subseteq V$.

Consider two disjoint open sets in $Y$, which will be of the form $U \cap Y$, $V \cap Y$. Then $Y$ is disconnected if and only if

- They are disjoint, i.e. $U \cap V \cap Y = \varnothing$
- They cover $Y$, i.e. $Y \subseteq U \cup V$
- Neither is empty

So for $Y$ to be connected, at least one of $U$ and $V$ must be empty, i.e. $Y \subseteq U$ and $Y \subseteq V$.

Can you state the sunflower lemma?

Let $\{A _ i \vert i \in I\}$ be a collection of connected subsets of $X$ such that

\[\bigcap_{i \in I} A_i \ne \varnothing\]then

\[\bigcup_{i \in I} A_i\]is connected.

Quickly explain the idea behind proving the sunflower lemma:

Let $\{A _ i \vert i \in I\}$ be a collection of connected subsets of $X$ such that

\[\bigcap _ {i \in I} A _ i \ne \varnothing\]
then

\[\bigcup _ {i \in I} A _ i\]
is connected.

Let $\{A _ i \vert i \in I\}$ be a collection of connected subsets of $X$ such that

Consider any continuous function on the union, since each $A _ i$ is connected, $f$ must be constant on each $A _ i$. Then since all the $A _ i$ overlap, this forces $f$ to be constant on the union also.

Can you state the lemma about a set being connected if it’s “between” a connected set and its closure?

Let $A \subseteq X$ be connected. Then for all $B$ such that

\[A \subseteq B \subseteq \bar A\]$B$ is connected.

What lemma relates continuous maps and connected sets?

Let $X$ be a connected metric space, and let $f : X \to Y$ be continuous. Then $f(X)$ is connected.

What is the connected component of $X$ containing some $x$, denoted $\Gamma(x)$?

The union of all connected subsets of $X$ containing $X$.

What notation is used for the connected component of $X$ containing $x$?

What are the connected components of $\mathbb R$?

The intervals.

What does it mean for a metric space to be totally disconnected?

Every connected component is just a single point.

Quickly prove that if:

- $(X, d _ X)$ is connected
- $(Y, d _ Y)$ is connected

then:

- $X \times Y$ is connected

- Choose some $(x, y) \in X \times Y$.
- Note that $X \times \{y\}$ is connected as it is homeomorphic to $X$ via $x \mapsto (x, y)$ and likewise $\{x\} \times Y$ is connected
- Define $T _ p := (X \times \{y\}) \cup (\{p\} \times Y)$, then $T _ p$ is connected since both sets have $(p, y)$ in common
- But then $X \times Y = \bigcup _ {p \in X} T _ p$, which is a union of connected subsets, and since they all contain $(x, y)$, it must also be connected.

Suppose we are considering $\mathbb Z \subseteq \mathbb R$ and want to show that each $\{n\}$ is a connected component of $\mathbb Z$. What’s an easy way to do this?

Note that $\{n\}$ is both open and closed.

Quickly prove that the following three characterisations of connectedness for a metric space $X$ are equivalent:

- $X$ is connected
- If $f : X \to [0, 1]$ is continuous, it is constant.
- The only open and closed subsets of $X$ are $X$ and $\varnothing$.

- $(1) \implies (2)$: Suppose $X$ is connected and $f : X \to \{0, 1\}$ is a continuous function. Then the singleton sets $\{0\}$ and $\{1\}$ are both open in $\{0, 1\}$ so both $f^{-1}(0)$ and $f^{-1}(1)$ are open in $X$. They are clearly disjoint, and their union is $X$. Therefore one of them is empty, so $f$ is constant.
- $(2) \implies (3)$: Suppose that $A \subseteq X$ is both open and closed. Then $A^C$ is also open, so the function $f : X \to \{0, 1\}$ defined by $f(x) = 1$ for $x \in A$ and $f(x) = 0$ for $x \notin A$ (i.e. $f(x) = \mathbb 1 _ A(x)$) is continuous (because of the preimage property). If it is just $1$, then $A = X$. If it is just $0$, then $A = \emptyset$.
- $(3) \implies (1)$: Suppose that $X = U \cup V$ with $U$, $V$ open and disjoint. Then $U^C = V$ is open, so $U$ is also closed. Thus $U$ is both open and closed, and hence is either $X$ or $\emptyset$. Similarly for $V$. Hence there is no way to disconnect $f(X)$.

What does it mean for a subset $J \subseteq \mathbb R$ to have the interval property?

For all $x, y \in J$ such that $x < y$, $[x, y] \subseteq J$.

Quickly prove that a subset $J \subseteq \mathbb R$ is an interval if and only if it is connected, assuming that the following holds:

$J$ is an interval $\iff$ for all $x, y \in J$, if $x < y$, then $[x, y] \subseteq J$ (“the interval property”).

$J$ is an interval $\iff$ for all $x, y \in J$, if $x < y$, then $[x, y] \subseteq J$ (“the interval property”).

**$J$ interval $\implies$ $J$ connected**: Consider an arbitrary continuous function $f : J \to \{0, 1\}$. Then $f$ can also be considered a continuous function $f : J \to [0, 1]$.

Assume for a contradiction that $f$ is not constant. Then $\exists x, y \in J$ such that $f(x) = 0$ and $f(y) = 1$. Wlog, we can assume $x < y$. Then since $J$ satisfies the interval property, $[x, y] \subseteq J$. Then applying the intermediate value theorem to the restriction of $f$ to $[x, y]$, $\exists \xi \in [x, y]$ such that $f(\xi) = 1/2$, a contradiction.

**$J$ connected $\implies$ $J$ interval**: Fix some $x, y \in J$ such that $x < y$. Want to show that $[x, y] \subseteq J$. Assume for a contradiction that $\exists c \in [x, y]$ but $c \notin J$. Then

Then $U, V$ open and disjoint, and $J = U \cup V$. So $J$ is disconnected, a contradiction.

Quickly prove that if:

- $X$ is a connected metric space
- $f : X \to Y$ is continuous

Then

- $f(X)$ is connected.

We can assume (by replacing $Y$ with $f(X)$) that $f$ is surjective. Then suppose $U, V$ are disjoint, open subsets of $Y$ such that

\[Y = U \cup V\]Then

\[X = f^{-1}(U) \cup f^{-1}(V)\]and $f^{-1}(U), f^{-1}(V)$ are disjoint and open. Since $X$ is connected, one of them must be empty. Wlog, say $f^{-1}(U)$ is empty. Then $U$ is empty, $Y$ cannot be disconnected.

Quickly prove that if $A \subseteq X$ is connected, then for all $B$ such that

\[A \subseteq B \subseteq \bar A\]
$B$ is connected.

We use the following lemma:

Let $X$ be a metric space and $Y \subseteq X$. Then $Y$ is connected if and only if the following is true: If $U, V$ are open subsets of $X$ and $U \cap V \cap Y = \varnothing$, then whenever $Y \subseteq U \cup V$, either $Y \subseteq U$ or $Y \subseteq V$.

Applying this to $B$, suppose $B \subseteq U \cup V$ where $U, V$ open in $X$ and $U \cap V \cap B = \emptyset$.

Then $A \subseteq U \cup V$ and $A \cap U \cap V = \emptyset$.

Since $A$ is connected, either $A \subseteq U$ or $A \subseteq V$.

Wlog, assume $A \subseteq U$. Since $A \cap U \cap V = \emptyset$, $A \subseteq V^C$.

But $V^C$ is closed, so taking closures,

\[\overline A \subseteq \overline{V^C} = V^C\]Then since $B \subseteq \overline A$, $B \subseteq V^C$. Then $B \subseteq U \cup V$ implies that $B \subseteq U$, so the conditions of the lemma are satisifed.