ANA1_L22

Connected sets

Definition 1(Definition).

For a metric space $X$, a separation of $X$ is a partition of $X=A\bigsqcup B$ into two proper subsets $A$ and $B=X\setminus A$ such that $A$ and $B$ are both open in $X$.

Note that if $A$ and $B$ are both open in $X$, it follows that $A$ and $B$ are both closed in $X$. A set which is both open and closed in called a clopen set. So, $A$ and $B$ are clopen in $X$.

Definition 2(Definition).

If $X$ has a separation, $X$ is called disconnected. If $X$ is not disconnected, $X$ is connected.

For example, $X=\mathbb{R}\setminus \{p\}$ is disconnected because $X=(-\infty ,p)\bigsqcup (p,\infty )$.

Characterization of connected sets in R

Theorem 3(Theorem).

A nonempty subset $X\subset \mathbb{R}$ is connected $⟺$ (($p,q\in X$ and $p<x<q$ )$⟹$ $x\in X$), i.e, a connected set in $\mathbb{R}$ is either a singleton, an interval, a ray, or $\mathbb{R}$.

Proof of $⟹$
We will prove the contrapositive. Let $p<x<q$ with $x\notin X$, $p,q\in X$. Let $A=X\cap (-\infty ,x)$ and $B=X\cap (x,\infty )$. $X=A\bigsqcup B$. ❏

Proof of $⟸$
Suppose $X=A\bigsqcup B$, with $A$ and $B$ nonempty, disjoint and open in $X$. Let $a\in A$, $b\in B$, and WLOG $a<b$. We know $[a,b]\subset X$. Let $c\equiv \inf [a,b]\cap B$. Clearly, $c\notin A$, since then every neighborhood of $c$ must contain a point of $B$, making it impossible to find a neighborhood of $c$ in $A$ (Alternatively, $B$ being a closed set forces $c\in B$). $c\in B$ also leads to a contradiction, since B must contain a neighborhood of $c$, which in turn would contain an element smaller than $c$. $\Rightarrow \Leftarrow$ ❏

Images of connected sets under continuous functions are connected

Theorem 4(Theorem).

If $X$ is a connected metric space and $f:X\to Y$ is continuous, $f(X)$ is connected.

Proof
Let $f(X)=A\bigsqcup B$, $A$, $B$ open in $f(X)$. This implies $X=f^{-1}(A)\bigsqcup f^{-1}(B)$. $\Rightarrow \Leftarrow$ ❏

Intermediate value theorem

Rudin, 4.23

Theorem 5(Theorem).

Let $f:[a,b]\to \mathbb{R}$ be continuous, WLOG $f(a)<c<f(b)$. Then, $\exists x\in (a,b)$ such that $f(x)=c$.

Proof
From the previous theorem, we know $f([a,b])$ is connected. From the characterization of connected sets in $\mathbb{R}$, we know that $c\in f([a,b])$. ❏

Important

If $f$ satisfies the conclusion of the intermediate value theorem ($\forall c$ such that $f(a)<c<f(b)$, $\exists x\in (a,b)$ such that $f(x)=c$), $f$ need not be continuous. For example, consider $f:[-\pi ,\pi ]\to \mathbb{R}$ defined by

$$f(x)=\{\begin{array}{ll}\sin \left(\frac{1}{x}\right)& x\in [-\pi ,\pi ]\setminus \{0\}\\ 0& x=0\end{array}.$$

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Separated sets

Definition 6(Definition).

Subsets $P$ and $Q$ of $X$ are called separated if $\overline{P}\cap Q=\varnothing$ and $P\cap \overline{Q}=\varnothing$, where the closures are taken in $X$.

Note that $\overline{P}\cap \overline{Q}$ need not be empty.

Theorem 7(Theorem).

$P$ and $Q$ are separated in $X$ $⟺$ $P\bigsqcup Q$ is a separation of $P\cup Q$.

Proof of $⟹$
Closure of $P$ in $P\cup Q$ = (Closure of $P$ in $X$) $\cap$ ($P\cup Q$), which is equal to $P$, since (Closure of $P$ in $X$) $\cap$ $Q$ = $\varnothing$. Thus, $P$ is closed in $P\cup Q$. Similarly, $Q$ is closed in $P\cup Q$. ❏

Proof of $⟸$
We have to show that no limit point of $P$ in $X$ is contained in $Q$ (and vice versa). FTSOC, let $x\in X$ be a limit point of $P$, and $x\in Q$. This implies $x\in P\cup Q$, which implies $x\in P$, since $P$ is closed in $P\cup Q$. But, $P$ and $Q$ are disjoint. $\Rightarrow \Leftarrow$
Similarly, no limit point of $Q$ in $X$ is contained in $P$. ❏

Theorem 8(Corollary).

$P$ and $Q$ are separated in $X$ $⟺$ $P$ and $Q$ are separated in $Z\supset X$.

Rudin’s definition of connectedness

Definition 9(Definition).

$X\subset Z$ is disconnected if $X$ is a union of two nonempty separated sets.
If $X$ is not disconnected, $X$ is connected.

From the previous theorem (take $X$ to be $P\cup Q$), this is equivalent to our definition in the beginning.

Some theorems

Theorem 10(Theorem 1).

If $A$ and $B$ form a separation of $X$ then any connected subset of $X$ is contained in $A$ or in $B$.

Proof
Let $C\subset X$ be connected. Let $A\bigsqcup B=X$ be a separation of $X$. FTSOC, assume $C\cap A\ne \varnothing$ and $C\cap B\ne \varnothing$. Since $A$ and $B$ are open in $X$, $C\cap A$ and $C\cap B$ are open in $C$. We can write $C=(C\cap A)\bigsqcup (C\cap B)$, i.e, $C$ is the disjoint union of two nonempty sets open in $C$. But this means $C$ is disconnected! $\Rightarrow \Leftarrow$ ❏

Theorem 11(Theorem 2).

If a family of connected subsets of $X$ have a point in common, then their union is connected.

Proof
Let $\{C_{\alpha }\}$ be a family of connected subsets of $X$, and let $p\in {\bigcap }_{\alpha }{C}_{\alpha }$. FTSOC, assume $C\equiv {\bigcup }_{\alpha }{C}_{\alpha }$ is disconnected, i.e, there exist non empty, disjoint $A$ and $B$ such that $A\bigsqcup B=C$ and $A$ and $B$ are open in $C$. WLOG, $p\in A$. Consider $q\in B$. There must exist $C_{\alpha }$ such that $p,q\in C_{\alpha }$. Since $A$ and $B$ are open in $C$, it follows that $A\cap C_{\alpha }$ and $B\cap C_{\alpha }$ are open in $C_{\alpha }$. We can write $C_{\alpha }=(A\cap C_{\alpha })\bigsqcup (B\cap C_{\alpha })$, i.e, $C_{\alpha }$ is the the disjoint union of two nonempty sets open in $C_{\alpha }$. But this means $C_{\alpha }$ is disconnected! $\Rightarrow \Leftarrow$ ❏

Theorem 12(Theorem 3).

Let $E\subset X$. $E$ is connected $⟹$ $\overline{E}$ is connected.

Proof
FTSOC, suppose $\overline{E}=A\bigsqcup B$, where $A$ and $B$ are disjoint nonempty open subsets of $\overline{E}$. Then $E=(A\cap E)\bigsqcup (B\cap E)$. Note that $(A\cap E)$ and $(B\cap E)$ are open in $E$ as $A$ and $B$ are open in $\overline{E}$. Since $E$ is connected, one of $(A\cap E)$ or $(B\cap E)$ must be empty. WLOG, suppose $(B\cap E)=\varnothing$, which implies $A\supset E$.
This means $B$ contains only points from $\overline{E}\setminus E$, and since $B$ is nonempty, it must contain at least one point from $\overline{E}\setminus E$. However, this implies $A$ does not contain all its limit points (as points in $B$ are limit points of $A$), contradicting the assumption that $A$ is closed. $\Rightarrow \Leftarrow$ ❏

Connected components

Define $x\sim y$ if points $x$ and $y$ belong to a connected subset of $X$. This is an equivalence relation on $X$:

• $\sim$ is reflexive, since every singleton set is connected.
• $\sim$ is symmetric.
• $\sim$ is transitive, because of theorem 2.

Theorem 13(Theorem 4).

The equivalence classes of $\sim$ are connected.

Proof
Pick any $x\in X$ and consider its equivalence class $C_x$. $x\sim y$ for all $y\in C_x$, i.e, $x,y\in S_y$ for some connected $S_y\subset X$ for all $y\in C_x$. Note that $S_y\subset C_x$ for all $y$. Thus, $C_x={\bigcup }_{y\in C_x}{S}_y$. Also note that $x\in {\bigcap }_{y\in C_x}{S}_y$. So, from theorem 2, $C_x$ is connected. ❏

The equivalence classes of $\sim$ are hence called connected components of $X$.

Theorem 14(Theorem 5).

Any connected subset of $X$ is contained in one of its connected components.

Proof
Let $E\subset X$ be connected, with $x,y\in E$ and $x\in C_1$ and $y\in C_2$, where $C_1$ and $C_2$ are connected components of $X$. Then, $x\nsim y$, implying $E$ is not connected. $\Rightarrow \Leftarrow$ ❏

Theorem 15(Theorem 6).

Connected components of $X$ are closed in $X$.

Proof
Let $C$ be a connected component of $X$. From theorem 3, $\overline{C}$ must be connected. From theorem 5, $\overline{C}$ must then be contained in a connected component of $X$. This forces $C=\overline{C}$. ❏

The components are also open in $X$ if their number is finite, since finite unions of closed sets are closed.

The connected components of $\mathbb{Q}$ are singletons, since for any interval $I=\mathbb{Q}\cap [a,b]$ one can find $c\notin \mathbb{Q}$ such that $I=(I\cap (a-1,c))\bigsqcup (I\cap (c,b+1))$. Such a space is called totally disconnected.