3.1Connectedness

Updated 6 Jul 2026Chapter (PDF)

[3.1.1]Definition(Separations and Connectedness)#

Let $X$ be a topological space. A separation of $X$ is a pair of nonempty, disjoint open subsets $U, V \subseteq X$ whose union is $X$ . We say $X$ is connected if there exists no separation of $X$ .

[3.1.2]Remark(Intuition Behind Separation)#

Intuitively, we could consider a topological space $X$ to be disconnected if we can find two non-empty, disjoint subsets $A, B \subseteq X$ whose union is $X$ , such that neither set contains a limit point of the other. The requirement that neither set possesses any of the other's limit points entails that $A$ and $B$ are topologically isolated from each other, incapable of touching even at their boundaries. This is the geometric idea behind a disconnection. If $X$ cannot be partitioned into sets that strictly exclude each other's limit points, then no part of $X$ can be cleanly detached from the rest, meaning the space is connected. The equivalence between this geometric idea and the framing of a disconnection in terms of open sets will be visited soon.

[3.1.3]Theorem#

Separations are formed by clopen sets.

Proof.

Suppose $X$ is a topological space, and suppose $A, B \subseteq X$ are disjoint, non-empty, open sets whose union is $X$ . Note also that $B = X \setminus A$ and $A = X \setminus B$ , which are closed sets, form a separation as well. Thus $A, B$ are clopen.

[3.1.4]Theorem#

A topological space $X$ is connected if and only if the only clopen subsets of $X$ are $\emptyset$ and $X$ .

Proof.

$(\Longrightarrow)$ : Suppose $X$ is connected. Then let $O$ be a clopen subset of $X$ . Note that $O, X \setminus O$ are disjoint open subsets of $X$ whose union is $X$ . For $X$ to be connected, it cannot be separated, meaning $O, X \setminus O$ cannot be non-empty. Thus the only clopen subsets of $X$ are $\emptyset, X$ .

$(\Longleftarrow)$ : Suppose the only clopen subsets of $X$ are $\emptyset, X$ . By the previous theorem, there exists no separation of $X$ , so $X$ is connected.

We reformulate the idea of connectedness in terms of limit points.

[3.1.5]Theorem#

Let $Y$ be a subspace of the topological space $X$ . A separation of $Y$ is defined to be a pair of disjoint, non-empty sets $A, B \subseteq Y$ such that neither contains the limit points of the other. Then $Y$ is connected if no separation exists.

Proof.

First, we show that disjoint, non-empty sets $A, B \subseteq Y$ such that neither contains the limit points of the other are open. Since $\Cl_Y(A) \cap B = A \cap \Cl_Y(B) = \emptyset$ , and $\Cl_Y(O) = \Cl_X(O) \cap Y$ for a general $O \subseteq Y$ , it follows that $\Cl_X(A) \cap B = A \cap \Cl_X(B) = \emptyset$ . Thus $\Cl_X(A) \cap Y = A$ and $\Cl_Y(B) \cap Y = B$ , meaning $A, B$ are closed in $Y$ . Thus $A, B$ are open as well (since $B = A^c$ and $A = B^c$ ).

Suppose $A, B \subseteq Y$ are disjoint, non-empty open sets whose union is $Y$ . Note that $A, B$ are clopen, and so $A = \Cl_Y(A) = \Cl_X(A) \cap Y$ , and the intersection of this with $B$ is $\emptyset$ . Analogously, $A \cap \Cl_Y(B) = \emptyset$ as desired.

[3.1.6]Theorem#

Suppose $X$ is a topological space with some separation $C, D \subseteq X$ . If a subspace $Y \subseteq X$ is connected, then $Y$ lies entirely within either $C$ or $D$ .

Proof.

Note separations are formed by open sets, so $C, D \subseteq X$ are open in $X$ and consequently $C \cap Y, D \cap Y \subseteq Y$ are open. Then

(C \cap Y) \cap (C \cap D) = (C \cap D) \cap Y = \emptyset \cap Y = \emptyset,

(C \cap Y) \cup (C \cap D) = (C \cup D) \cap Y = X \cap Y = Y.

$Y$ is connected and thus no separation for it exists, but $C \cap Y, D \cap Y$ would form a separation for $Y$ if one is non-empty. Thus, $Y = C \cap Y$ or $Y = D \cap Y$ , meaning $Y$ is a subset of one of $C, D$ .

[3.1.7]Theorem(Arbitrary Union of Connected Spaces are Connected)#

Suppose $X$ is a topological space and $\{O_\lambda\}_{\lambda \in \Lambda}$ is a family of connected subsets of $X$ , each containing some common point $p \in X$ . Then $O = \bigcup_{\lambda \in \Lambda} O_\lambda$ is connected.

Proof.

Suppose $X$ is a topological space and $\{O_\lambda\}_{\lambda \in \Lambda}$ is a family of connected subsets of $X$ , each containing some common point $p \in X$ . Let $O = \bigcup_{\lambda \in \Lambda} O_\lambda$ . For the sake of contradiction, suppose $A, B \subseteq O$ is a separation for $O$ . Without loss of generality, suppose $p \in A$ . Then note each $O_\lambda$ is connected, and so $O_\lambda \subseteq A$ is forced ( $O_\lambda \not \subseteq B$ since $p \notin B$ ). Thus $O \subseteq A$ , meaning $B = \emptyset$ , a contradiction.

[3.1.8]Theorem(Continuous Image of Connected Set is Connected)#

The continuous image of a connected set is connected.

Proof.

Let $f: X \to Y$ be a continous map from a connected topological space $X$ to some set $Y$ . Note then that $\tilde f: X \to f(X)$ is continuous—we seek to show $f(X)$ is connected. For the sake of contradiction, suppose $A, B \subseteq f(X)$ forms a separation for $f(X)$ . Note that

X = \tilde{f}^{-1}(f(X)) = \tilde{f}^{-1}(A \cup B) = \tilde{f}^{-1}(A) \cup \tilde{f}^{-1}(B),

where the last expression is the union of two open sets by continuity of $\tilde f$ . Then this is a separation of $X$ , but $X$ is connected, which is a contradiction.

An immediate corollary to this theorem is the Intermediate Value Theorem, provided we have proven connected subsets of $\bR$ are equivalent to intervals. We prove this later.

[3.1.9]Theorem(Finite Products of Connected Sets are Connected)#

A finite product of connected topological spaces is connected.

Proof.

Let $X, Y$ be connected topological spaces: we seek to show $X \times Y$ is connected as well. To start, fix $(a, b) \in X \times Y$ . Then note

X \cong (X \times b) \quad Y \cong (a \times Y).

Since connectedness is a topological property, it is preserved under homeomorphism. Thus

T_{(a, b)} = (X \times b) \cup (a \times Y)

is a union of two connected subspaces of $X \times Y$ that contains the point $(a, b)$ , so it is also connected. Then note

X \times Y = \bigcup_{(a, b) \in X \times Y} T_{(a, b)},

and so $X \times Y$ is connected. We may generalize this to finitely many products by induction.

We will prove later that $\bR$ is connected, but it is pretty intuitive to understand why. We discuss another advantage of the product topology over the box topology, in the fact that the product topology preserves more topological properties than the box topology.

[3.1.10]Remark(Connectedness of

\bR^\omega

Under Box and Product Topologies)#

Consider $\bR^\omega$ endowed with the box topology. Then note the set of bounded sequences, say $O$ , and unbounded sequences, say $O^c$ , form a partition for $\bR^\omega$ . We simply show these sets are open in the box topology. For any $\vb{a} \in \bR^\omega$ , define

U_a = (a_1 - 1, a_1 + 1) \times (a_2 - 1, a_2 + 1) \times \cdots,

which is open in the box topology. Then note each $U_{\vb{a}} \subseteq O$ for a bounded sequence $\vb{a}$ , and so $O = \bigcup_{\vb{a} \in O} U_{\vb{a}}$ , and so $O$ is open. $O^c$ can be proven to be open similarly, so $O, O^c$ form a separation for $\bR^\omega$ under the box topology.

Now consider $\bR^\omega$ endowed with the product topology. For each $n \in \bZ_+$ , define $\tilde \bR^n$ as the subspace of $\bR^\omega$ that is the set of all sequences $\vb{x} = (x_1, x_2, \dots)$ , where each $x_i = 0$ for every $i > n$ . Of course, $\tilde \bR^n \cong \bR^n$ for every $n \in \bZ_+$ , so these subspaces are connected. We define $\bR^\infty = \bigcup_{n \in \bZ_+} \tilde \bR^n$ , which is connected since each space share the common zero sequence. From an intuitive standpoint, we recognize $\bR^\infty$ as the set of all sequences whose tails zero out. We seek to show that $\bR^\omega$ , endowed with the product topology, is the topological closure for $\bR^\infty$ —since the closure of a connected set is connected, our work will be done. For any $\vb{a} \in \bR^\omega$ , we show that any open neighborhood about $\vb{a}$ intersects $\bR^\infty$ . Let $O$ be a basic open set of $\bR^\omega$ that contains $\vb{a}$ . In the product topology, this means there is some $n \in \bZ_+$ such that

O = \left( \prod_{k = 1}^n (a_k - \epsilon_k, a_k + \epsilon_k) \right) \times \prod_{n \in \bZ_+ \setminus \{1, \cdots, n\}} \bR.

Thus $(a_1, a_2, \dots, a_n, 0, \dots) \in O$ , so $O \cap \bR^\infty \ne \emptyset$ .

Even though $\bR$ is connected, the box topology does not preserve connectedness when $\bR$ is multiplied by itself countably many times, whereas the product topology does. In fact, we prove later that an arbitrary product of connected sets is connected under the product topology.

3.1.1Exercises#

[3.1.11]Problem#

Let $X$ be a topological space, and let $\{A_n\}_{n \in \bZ_+}$ be a family of connected topological subsets of $X$ such that $A_n \cap A_{n + 1} \ne \emptyset$ . Show $A = \bigcup_{n \in \bZ_+} A_n$ is a connected subset of $X$ .

Solution.

For the sake of contradiction, suppose $A = B \cup C$ is a separation. Recall that each $A_n$ is a connected subset of $X$ : if $A_n$ were disconnected in $A$ , then a separation of it would exist in $A$ (and thus consequently $X$ ), so $A_n$ is also a connected subset of $A$ in its subspace topology. Without loss of generality, we can say $A_1 \subseteq B$ . Then $A_2 \subseteq B$ or $A_2 \subseteq C$ , but $A_1 \cap A_2 \ne \emptyset$ and so $A_2 \subseteq B$ is forced. By induction, it follows every $A_n \subseteq B$ , and so $A \subseteq B$ and $C = \emptyset$ , a contradiction.

[3.1.12]Problem#

Let $A \subseteq X$ . Suppose $C \subseteq X$ is a connected subspace of $X$ that intersects $A$ and $X \setminus A$ . Show that $C \cap \partial A \ne \emptyset$ .

Proof.

Note that since $C \cap A \ne \emptyset$ and $C \cap A^c \ne \emptyset$ , it follows that $C \cap \bar{A} \ne \emptyset$ and $C \cap \bar{(X \setminus A)} \ne \emptyset$ . Moreover, note

C = C \cap X = C \cap (\bar{A} \cup \bar{(X \setminus A)}) = \underbrace{(C \cap \bar{A})}_{:= A} \cup \underbrace{(C \cap \bar{(X \setminus A)})}_{:= B}.

If $A \cap B = \emptyset$ , then each set contains neither of the other's limit points and so this would be a separation of $C$ , which is impossible since $C$ is connected. Thus there is some $p \in A \cap B$ , implying $p \in C \cap \partial A$ .

[3.1.13]Problem#

Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a family of connected spaces, and let $X = \prod_{\lambda \in \Lambda} X_\lambda$ . Fix $a = (a_\lambda)_{\lambda \in \Lambda}$ .

For any finite $K \subseteq \Lambda$ , define $X_K$ to be the set of all points $x = (x_\lambda)_{\lambda \in \Lambda}$ such that $x_\lambda = a_\lambda$ for every $\lambda \notin K$ . Show $X_K$ is connected.
Show that $Y = \bigcup_{K \subseteq \Lambda, |K| < \infty} X_K$ is connected.
Show that $X = \bar{Y}$ and conclude that $X$ is connected.

This result shows that the arbitrary product of connected sets is connected as well.

Proof.

Note that
$X_K \cong \prod_{\lambda \in K} X_\lambda \times \prod_{\lambda \notin K} \{a_\lambda\} = C \times \prod_{\lambda \notin K} \{a_\lambda\},$
where $C$ is the product of finitely many connected sets. Moreover, note $\prod_{\lambda \notin K} \{a_\lambda\}$ is a singleton, which is trivially connected. Thus, the product of these two yields a connected subset of $X$ , and since it is homeomorphic to $X_K$ , we are done.
Note each $X_K$ , for finite $K$ , is connected. Moreover, each shares the common point $a$ , so their union is connected.
Note that $Y \subseteq X$ , so $\bar{Y} \subseteq X$ immediately. Thus we show every $X \subseteq \bar{Y}$ . Let $x \in X$ , and let $O$ be a basic open neigbhorhood of $x$ in the product topology of $X$ . Then note $O = \prod_{\lambda \in \Lambda} U_\lambda$ , where $U_\lambda = X_\lambda$ for all but finitely many valued of $\lambda$ , which we say are $\lambda_1, \dots, \lambda_n$ . Thus there is some point in $O$ with arbitrary terms for indices $\lambda_1, \dots, \lambda_n$ (such that they satisfy the product topology condition) and $a_\lambda$ for all other values of $\lambda$ . This point also belongs to $Y$ , so $O \cap Y \ne \emptyset$ . Since this basic open set is arbitrary, it follows that $X \subseteq \bar{Y}$ , completing the proof.