Notes - Metric Spaces MT23, Connectedness

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

• Nonempty
• Open
• Disjoint

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

\[f(x) = \mathbb 1_A\]

or

\[f(x) = \mathbb 1_B = 1-\mathbb 1_A\]

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 $Y \subseteq V$

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 $Y \subseteq V$.

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.

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.

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

\[A \subseteq B \subseteq \bar A\]

$B$ is connected.

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

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

\[\Gamma(x)\]

The intervals.

Every connected component is just a single point.

• 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.

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

• $(1) \implies (2)$: 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 $f(X)$.

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

$J$ 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

\[\begin{aligned} U &= J \cap (-\infty, c) \\\\ V &= J \cap (c, \infty) \end{aligned}\]

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

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.

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.