Connected and Disconnected Topological Spaces Review

# Connected and Disconnected Topological Spaces Review

We will now review some of the recent material regarding connected and disconnected topological spaces.

- On the
**Connected and Disconnected Topological Spaces**page we said that a topological space $X$ is said to be**Disconnected**if there exists open sets $A, B \subset X$ such that $A, B \neq \emptyset$, $A \cap B = \emptyset$ and $X = A \cup B$. In other words, a topological space is disconnected if there exists nonempty, proper and disjoint open sets $A$ and $B$ whose union is $X$. If such sets exist then $\{ A, B \}$ is called a**Separation**of $X$.

- Furthermore, we said that a topological space that is not disconnected is said to be
**Connected**.

- We saw that the set of rational numbers $\mathbb{Q}$ with the subspace topology from $\mathbb{R}$ (with the usual topology) is disconnected since if we consider the rational number $\sqrt{2}$ and let $A = \{ q \in \mathbb{Q} : q < \sqrt{2} \}$ and $B = \{ q \in \mathbb{Q} : q > \sqrt{2} \}$ then $\{ A, B \}$ is a separation of $\mathbb{Q}$.

- On the
**Connected and Disconnected Sets in Topological Spaces**we defined connectivity for sets in a topological space. We said that a set $A \subseteq X$ is connected if $A$ equipped with the subspace topology from $X$ is a connected topological space. Similarly, we said that $A$ is disconnected if $A$ equipped with the subspace topology from $X$ is a disconnected topological space.

- We looked at the example of the set $A \subset \mathbb{R}$ defined by $A = (0, 1) \cup [2, 3]$ as an example of a disconnected set in $\mathbb{R}$. A

- On the
**Disconnected Topological Spaces Homeomorphic to the Topological Sum of their Separation**page we looked at a nice characterization of disconnected topological spaces. We proved that $X$ is a disconnected topological space with separation $\{ A, B \}$ if and only if $X$ is homeomorphic to the topological sum $A \oplus B$ where $A \oplus B$ is the set $A \cup B$ whose topology is given by the basis:

\begin{align} \quad \tau = \{ U \subseteq A \cup B : U \: \mathrm{is \: open \: in \: A} \: \mathrm{or} \: U \: \mathrm{is \: open \: in \: B} \} \end{align}

- On the
**Continuous Two-Valued Function Criterion for Disconnected Topological Space**page we looked at another characterization of disconnected topological spaces. We saw that $X$ is disconnected if and only if there exists a continuous and surjective function $f : X \to \{ 0, 1 \}$ (where $\{ 0, 1 \}$ has the discrete topology). We accomplished this by taking any separation $\{ A, B \}$ of $X$ and letting the image of $A$ under $f$ be $\{ 0 \}$, and letting the image of $B$ under $f$ be $\{ 1 \}$, and conversely, if such a function exists then we can take $\{ f^{-1}(0), f^{-1}(1) \}$ to be a separation of $X$.

- On the
**Clopen Set Criterion for Disconnected Topological Spaces**we looked at another nice characterization for disconnected/connected topological spaces. We saw that a topological space $X$ is disconnected if and only if $X$ contains a nonempty proper clopen (open and closed) subset. If such a subset $A \subset X$ exists then we can always form a separation of $X$ to be $\{ A, A^c \}$.

- Equivalently, a topological space $X$ is connected if and only if the empty set $\emptyset$ and whole set $X$ are the only clopen subsets of $X$.

- On
**The Connectedness of the Closure of a Set**page we said that if $A \subseteq X$ and if $A$ is a connected in $X$ then the closure $\overline{A}$ is also connected in $X$.

- We then began to consider whether the union of connected sets were connected or not. We looked at two very important theorems on this topic.

- Firstly, on the
**Common Point Criterion for Connectedness of Unions of Topological Subspaces**page we saw that if $\{ A_i \}_{i in I}$ is an arbitrary collection of topological subspaces of $X$ such that $\displaystyle{\bigcap_{i \in I} A_i \neq \emptyset}$, i.e., all of these sets share a common point, then $\displaystyle{\bigcup_{i \in I} A_i}$ is connected.

- Secondly, on the
**Common Connector Criterion for Connectedness of Unions of Topological Subspaces**page we said that if $\{ A_i \}_{i \in I}$ is an arbitrary collection of topological subspaces of $X$ ad $A_0$ is another connected subspace of $X$ such that $A_i \cap A_0 \neq \emptyset$ for all $i \in I$ then $\displaystyle{A_0 \cup \bigcup_{i \in I} A_i}$ is also connected.

- On the
**Preservation of Connectivity under Continuous Maps**page we saw that if $X$ and $Y$ are topological spaces, $f : X \to Y$, and $X$ is connected (in $X$), then the range, $f(X)$, is connected in $Y$

- We then looked at a nice criterion for determining whether two spaces are no homeomorphic. On the
**Connected/Disconnected Criterion for Non-Homeomorphic Topological Spaces**page we saw that if $X$ and $Y$ are topological spaces such that for every $x \in X$, $X \setminus \{ x \}$ is connected in $X$ and if there exists a $y \in Y$ such that $Y \setminus \{ y \}$ is disconnected in $Y$ then $X$ CANNOT be homeomorphic to $Y$.

- On the
**Non-Homeomorphic Topological Spaces Classified by Connectivity**page we looked at a good example of this result. We noted that the interval $(0, 1)$ is not homeomorphic to the unit circle in $\mathbb{R}^2$ since the removal of any point $x$ in $(0, 1)$ makes $(0, 1) \setminus \{ x \}$ disconnected, while the removal of any point on the unit circle leaves the unit circle connected.

- We then began to look at the connectivity of topological products.

- On the
**Connectedness of Finite Topological Products**page we proved that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of connected topological spaces then the topological product $\displaystyle{\prod_{i \in I} X_i}$ (with the product topology) is also connected. We noted that the result is also true if we consider the topological product of an arbitrary collection of connected topological spaces.

- On the
**Connectedness of Box Topological Products**we saw that if $\{ X_i \}_{i \in I}$ is an infinite collection of connected topological spaces then the box topological product $\displaystyle{\prod_{i \in I}^{\mathrm{Box}} X_i}$ might not be connected surprisingly enough!