Connected spaces

Separations

1. Separation

   Let $X$ be a topological space. A separation of $X$ is a pair of open sets $U,V$ such that:

   • $U\cap V=\emptyset$,
   • $U\cup V=X$,
   • $U\ne\emptyset$, $V\ne\emptyset$

2. Connected space

   A topological space is connected if it has no separation.

Remark

1. A space is connected if an only if the only subsets that are both open and closed are $\emptyset$ and $X$.

2. This is because if $U,V$ form a separation, then both are open and closed. Conversely, if $A\subseteq X$ is not the empty set and not $X$, and is open and closed, then $A,X-A$ form a separation.

Lemmas

1. :B_lemma:

   Let $X$ be a space and $Y$ a subspace of $X$. Then a separation of $Y$ is a pair of disjoint nonempty sets $A,B$ with union $Y$, and none containing a limit point of the other.

2. • Let $A,B$ be a separation of $Y$. Then both $A,B$ are open and closed in $Y$. We have that the closure of $A$ in $Y$ is $\overline{A}\cap Y$, and since $A$ is closed, we have $A=\overline{A}\cap Y$. Since $A,B$ are disjoint, we have $\overline{A}\cap B=\emptyset$.
   • Now, suppose $A,B\subseteq Y$ are disjoint, nonempty, and none containing a limit point of the other. Then $\overline{A}\cap B=\emptyset$, hence $\overline{A}\cap Y=A$. It follows that $A$ is closed in $Y$, hence $B$ is open in $Y$.

Examples

• If $X$ has the indiscrete topology, then $X$ is connected.
• If $X$ is a space with only one point, then it is connected.
• The subspace $Y=[-1,0)\cup (0,1]$ of $\mathbb{R}$ is not connected.
• $\mathbb{Q}$ as a subspace of $\mathbb{R}$ is not connected.

Properties of connected spaces

1. :B_lemma:

   If $A,B$ form a separation of $X$, and $Y\subseteq X$ is connected, then either $Y\subseteq A$ or $Y\subseteq B$.

2. We have that $Y\cap A,Y\cap B$ are both open in $Y$, they are disjoint and its union is $Y$. Since $Y$ is connected they cannot be both nonempty, and if $Y\cap A=\emptyset$, then $Y\subseteq B$.

3. :B_lemma:

   The union of connected subspaces with a point in common is connected.

4. Let $X=\cup_{\alpha\in I}A_{\alpha}$ with $A_{\alpha}$ connected, and $x_{0}\in\cap A_{\alpha}$. Let $U,V$ be a separation of $X$. Suppose $x_{0}\in U$. By the previous lemma, we must have $A_{\alpha}\subseteq U$ for all $\alpha\in I$. But then $V=\emptyset$, which is a contradiction.

More theorems

1. Teorema

   Let $A\subseteq X$ be connected. Then if $B\subseteq X$ is such that $A\subseteq B\subseteq\overline{A}$, then $B$ is also connected.

2. Proof

   Suppose that $C,D$ is a separation of $B$. By a previous lemma, we may assume without loss that $A\subseteq C$. But then $\overline{A}\subseteq \overline{C}$, and, since $\overline{C}$ and $D$ are disjoint and $B\subseteq \overline{A}$, we must have $B\cap D=\emptyset$, which is a contradiction. $\square{}$

3. Teorema

   Let $f\colon X\to Y$ be a continuous map, with $X$ connected. Then $f(X)$ is connected.

4. Proof

   Let $Z=f(X)$, and consider the surjective map $f\colon X\to Z$, which is also continuous. Let $A,B$ be a separation of $Z$. Then $f^{-1}(A),f^{-1}(B)$ would form a separation of $X$, which is impossible. Therefore, there is no separation of $Z$, hence $f(X)$ is connected. $\square{}$

5. Teorema

   The arbitrary product of connected spaces is connected.

6. • For the proof, we first prove that if $X,Y$ are connected, then $X\times Y$ is connected.
   • Let $(a,b)\in X\times Y$. Then $X\times\{b\}$ and $\{x\}\times Y$ are connected, for all $y\in Y$.
   • Hence $T_{x}=(X\times\{b\})\cup (\{x\}\times Y)$ is connected, since it is the union of two connected spaces with a point in common.
   • Then, $X\times Y=\cup_{x\in X}T_{x}$ is the union of connected spaces with the point $(a,b)$ in common.
   • By induction, we obtain that the product of any finite number of connected spaces is connected.

• Now, let $X_{\alpha}$ be a connected space, for $\alpha\in I$. We want to prove that $\prod_{\alpha}X_{\alpha}$ is connected. Choose $b=(b_{\alpha})\in X$.
• Given $\{\alpha_{1},\ldots,\alpha_{n}\}\subseteq I$, we define $X(\alpha_{1},\ldots,\alpha_{n})$ as the subspace of $X$ with points such that $x_{\alpha}=b_{\alpha}$ for $\alpha\not\in\{\alpha_{1},\ldots,\alpha_{n}\}$.
• We have that $X(\alpha_{1},\ldots,\alpha_{n})$ is homeomorphic to $X_{\alpha_{1}}\times\cdots\times X_{\alpha_{n}}$, and so it is connected.
• Let $Y$ be the subspace of $X$ that is the union of all $X(\alpha_{1},\ldots,\alpha_{n})$, for $\{\alpha_{1},\ldots,\alpha_{n}\}\subseteq I$ finite. Then $Y$ is connected, as is the union of connected spaces with the point $b$ in common.
• We finally prove that $\overline{Y}=X$. Let $x=(x_{\alpha})\in X$, and $U=\prod U_{\alpha}$ be a basis element of the product topology. We have that $U=X_{\alpha}$ for all $\alpha\in I-\{\alpha_{1},\ldots,\alpha_{n}\}$.
• Let $y=(y_{\alpha})$ defined as $y_{\alpha}=x_{\alpha}$ for $\alpha\in\{\alpha_{1},\ldots,\alpha_{n}\}$, and $y_{\alpha}=b_{\alpha}$ otherwise. Then $y\in X(\alpha_{1},\ldots,\alpha_{n})$, so $Y\cap U\neq\emptyset$. $\square{}$

Connected subsets of $\mathbb{R}$

Linear continuum

1. :B_definition:

   A totally ordered set $L$ is called a linear continuum if it satisfies the following:

   • $L$ has the least upper bound property.
   • If $x<y$, there is $z$ such that $x<z<y$.

2. Teorema

   If $L$ is a linear continuum, then $L$ is connected in the order topology. (Hence, every interval and every ray in $L$ is connected.)

Proof

• We say that a subset $Y\subseteq L$ is convex if $a,b\in Y$ with $a<b$ implies $[a,b]\subseteq Y$. We will prove that any convex subset $Y$ of $L$ is connected.
• Suppose that $Y=A\cup B$, with $A,B$ open in $Y$, disjoint and nonempty. Let $a\in A$, $b\in B$, and assume that $a<b$.

• Since $Y$ is convex, we have $[a,b]\subseteq Y$. Then $[a,b]$ is the union of the disjoint nonempty sets $A_{0},B_{0}$ given by:

  $$A_{0}=A\cap [a,b]\qquad B_{0}=B\cap [a,b]$$
• Note that $A_{0}$, $B_{0}$ are open in $[a,b]$, and thus form a separation of $[a,b]$.
• The interval $[a,b]$ is a subspace of $L$. Therefore the topology on $[a,b]$ is the order topology.
• Let $c=\sup A_{0}$. Then $c\in [a,b]$.

Proof (continuation)

• Suppose $c\in B_{0}$. Given that $c\ne a$, we have that either $c=b$ or $c\in (a,b)$.
• In any case, we have that there is $d\in [a,b]$ such that $(d,c]\subseteq B_{0}$.
• Hence $d$ is an upper bound of $A_{0}$, which contradicts the choice of $c$.
• Then $c\in A_{0}$. Given that $c\ne b$, we have that either $c=a$ or $c\in (a,b)$.
• In any case, we have that there is $d$ such that $[c,d)\subseteq A_{0}$.
• By Property 2 of linear continuum, there is $z$ such that $c<z<d$. But this contradicts that $c$ is upper bound of $A_{0}$. \qed

Connected subsets of $\mathbb{R}$

1. Corolario

   $\mathbb{R}$ is connected, and so is every interval and ray in $\mathbb{R}$.

2. Teorema

   If $X$ and $Y$ are linear continuum, and $Y$ is bounded above and below, then $X\times Y$ is a linear continuum with dictionary order.

Intermediate value theorem

1. Teorema

   Let $f\colon X\to Y$ be a continuous map, where $X$ is connected and $Y$ is ordered and has the order topology. Suppose that $a,b\in X$ and $y\in Y$ are such that $f(a)<y<f(b)$. Then there is $c\in X$ such that $f(c)=y$.

Path connectedness

1. :B_definition:
   • A path on a space $X$ from $x\in X$ to $y\in X$ is a continuous function $f\colon [a,b]\to X$ from some interval $[a,b]\subseteq \mathbb{R}$, such that $f(a)=x, f(b)=y$.
   • A space $X$ is path-connected if for every $x,y\in X$ there is a path from $x$ to $y$.

2. Teorema

   If $X$ is path-connected, then it is connected.

3. Examples

   The converse is not true, see $[0,1]\times [0,1]$ with dictionary order and the deleted comb space.