General Topology: Topological Spaces

Topological Spaces

A topology on a set $X$ is a collection $\mathcal{T}$ of subsets of $X$ with the following properties:

  1. $\varnothing$ and $X$ are in $\mathcal{T}$.
  2. For any subcollection $U$ of $\mathcal{T}$, the union of the sets in $U$ is also in $\mathcal{T}$.
  3. For any finite subcollection $U$ of $\mathcal{T}$, the intersection of the sets in $U$ is also in $\mathcal{T}$.

The set $X$ and a topology $\mathcal{T}$ together form an ordered pair $(X,\mathcal{T})$ called a topological space. Each element $U \in \mathcal{T}$ is called an open set of $X$.

We can see that two topologies on any set immediately present themselves. The first topology is the discrete topology, which is the topology $\mathcal{T}$ that consists of all subsets of $X$. The second is called the indiscrete topology, otherwise known as the trivial topology, which is the topology $\mathcal{T}$ whose only elements are $\varnothing$ and $X$.

Problems

  1. Let $X = \{1, 2, 3\}$. List all the possible topologies on $X$.

    1. $\{\varnothing, X\}$

    2. $\{\varnothing, \{1\}, X\}$

    3. $\{\varnothing, \{2\}, X\}$

    4. $\{\varnothing, \{3\}, X\}$

    5. $\{\varnothing, \{1, 2\}, X\}$

    6. $\{\varnothing, \{1, 3\}, X\}$

    7. $\{\varnothing, \{2, 3\}, X\}$

    8. $\{\varnothing, \{1\}, \{1, 2\}, X\}$

    9. $\{\varnothing, \{1\}, \{1, 3\}, X\}$

    10. $\{\varnothing, \{2\}, \{1, 2\}, X\}$

    11. $\{\varnothing, \{2\}, \{2, 3\}, X\}$

    12. $\{\varnothing, \{3\}, \{1, 3\}, X\}$

    13. $\{\varnothing, \{3\}, \{2, 3\}, X\}$

    14. $\{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, X\}$

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  2. Prove that the trivial topology is a topology.

    In order for a collection to be a topology, it must satisfy the three properties of topologies. For any set $X$, the discrete topology $\mathcal{T}$ is simply defined as $\mathcal{T} = \{ \varnothing, X \}$. 

    1. By definition, $\varnothing \in \mathcal{T}$ and $X \in \mathcal{T}$.

    2. The only possible union is the one between the only two elements of $\mathcal{T}$, and clearly $\varnothing \cup X = X$.

    3. The only possible intersection is likewise the one between the only two elements of $\mathcal{T}$, and clearly $\varnothing \cap X = \varnothing \in \mathcal{T}$.

    Having diligently checked all three conditions, we can confirm that the trivial topology is indeed a topology. Given the somniferous nature of this proof, we can also conclude that it was aptly named.

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  3. Prove that the discrete topology is a topology.

    Let $X$ be a set, and let $\mathcal{T}$ be the collection of all subsets of $X$. We check that $\mathcal{T}$ satisfies the three properties of a topology.

    1. By definition, $\varnothing \subseteq {X}$ and $X \subseteq {X}$, so $\varnothing \in \mathcal{T}$ and $X \in \mathcal{T}$.

    2. Let $\{U_{\alpha}\}$ be an indexed collection of subsets of $\mathcal{T}$. Because the elements of each $U_{\alpha}$ are all elements of $X$, then their union $\cup\{U_{\alpha}\}$ is also therefore only made up of elements of $X$, thus making it a subset of $X$ and therefore an element of $\mathcal{T}$.

    3. Let $\{U_1, \ldots, U_n \}$ be a finite subset of $\mathcal{T}$. Because the elements of each $U_{\alpha}$ are all elements of $X$, then their intersection $\cap\{U_{i}\}$ is also therefore only made up of elements of $X$, thus making it a subset of $X$ and therefore an element of $\mathcal{T}$.

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  4. Define the finite complement topology on a set $X$ as the collection $\mathcal{T}$ of all subsets of $X$ whose complement is either finite or all of $X$. Prove that the finite complement topology is a topology.

    1. $X^c = \varnothing$, and $\varnothing$ is finite, therefore $X \in \mathcal{T}$. Likewise, $\varnothing^c = X$, and $X$ is indeed $X$, so therefore $\varnothing \in \mathcal{T}$.

    2. Let $\{U_{\alpha}\}$ be an indexed subcollection of nonempty subsets of $X$. To show that $\cup \{U_{\alpha}\} \in \mathcal{T}$, we must show that $\left(\cup \{U_{\alpha}\}\right)^c$ is finite.

      $ \left(\cup U_{\alpha}\right)^c = X - \cup U_{\alpha} \\ \left(\cup U_{\alpha}\right)^c = \cap \left(X - U_{\alpha}\right) \\ $

      By definition, $X - U$ is finite for any $U \in \mathcal{T}$. Because the arbitrary intersection of finite sets is itself finite, it follows that $\left(\cup U_{\alpha}\right)^c$ is also finite and therefore also in $\mathcal{T}$.

    3. Let $\{U_1, \ldots, U_n\}$ be a finite collection of elements of $\mathcal{T}$. To show that $U_i \cap \ldots \cap U_n \in \mathcal{T}$, we must show that $\left(U_i \cap \ldots \cap U_n\right)^c$ is finite.

      $ \left(\bigcap\limits_{i=1}^{n} U_{i} \right)^c = X - \bigcap\limits_{i=1}^{n} U_{i} \\ \left(\bigcap\limits_{i=1}^{n} U_{i} \right)^c = \bigcup\limits_{i=1}^{n} \left(X - U_{i}\right) \\ $

      By definition, $X - U$ is finite for any $U \in \mathcal{T}$. Because the finite union of finite sets is itself finite, it follows that $\left(\bigcap\limits_{i=1}^{n} U_{i}\right)^c$ is also finite and therefore also in $\mathcal{T}$.

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  5. Let $X$ be a topological space, and let $A \subseteq X$ such that for each $x \in A$, there exists an open set $B$ such that $x \in B \subseteq A$. Show that $A$ is an open set.

    For all $x \in A$, there exists an open set $B_x \subseteq A$ that contains $x$. Therefore $A = \bigcup B_x$. Since $A$ is a union of open sets, it is itself open. 

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  6. Let $X$ be a set, and let $\{\mathcal{T_{\alpha}}\}$ be a collection of topologies on $X$. Determine whether $\bigcap \{\mathcal{T}_{\alpha}\}$ is a topology.

    To check whether $\bigcap \{\mathcal{T}_{\alpha}\}$ is a topology, we check the four requirements of a topology.

    1. Because each $\mathcal{T}_{\alpha}$ is a topology, it contains $\varnothing$ as an element. Therefore $\varnothing \in \bigcap \{\mathcal{T}_{\alpha}\}$.

    2. Likewise, because each $\mathcal{T}_{\alpha}$ is a topology, it contains $X$ as an element. Therefore $X\in \bigcap \{\mathcal{T}_{\alpha}\}$.

    3. Let $A_1, \ldots, A_n \in \bigcap \{\mathcal{T}_{\alpha}\}$. Then each $A_i$ is an element of every $\mathcal{T}_{\alpha}$. Because $A_i$ is open in $\mathcal{T}_{\alpha}$, it follows that $\bigcap\limits_{i = 1}^{n} A_i \in \mathcal{T}_{\alpha}$. Since this is true for all $\mathcal{T}_{\alpha}$, $\bigcap\limits_{i = 1}^{n} A_i \in \bigcap \{\mathcal{T}_{\alpha}\}$.

    4. Let $\{A_{\beta}\}$ be an indexed collection of open sets in $\bigcap\{\mathcal{T}_{\alpha}\}$. Then each $A_{\beta}$ is an element of every $\mathcal{T}_{\alpha}$. Because $A_{\beta}$ is open in $\mathcal{T}_{\alpha}$, it follows that $\bigcup \{A_{\beta}\} \in \mathcal{T}_{\alpha}$. Since this is true for all $\mathcal{T}_{\alpha}$, $\bigcup \{A_{\beta}\} \in \bigcap \{\mathcal{T}_{\alpha}\}$.

    Thus we conclude that $\bigcap \{\mathcal{T}_{\alpha}\}$ is a topology.

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  7. Let $X$ be a set, and let $\{\mathcal{T_{\alpha}}\}$ be a collection of topologies on $X$. Determine whether $\bigcup \{\mathcal{T}_{\alpha}\}$ is a topology.

    To check whether $\bigcup \{\mathcal{T}_{\alpha}\}$ is a topology, we check the four requirements of a topology. Alternatively, if it is not a topology, we need only find a counterexample.

    Let $X = \{1, 2, 3\}$, and let $\mathcal{T}_{\beta} = \{\varnothing, \{1\}, X\}$ and $\mathcal{T}_{\gamma} = \{\varnothing, \{2\}, \{2, 3\}, X\}$ be topologies on $X$. Then $\mathcal{T}_{\beta} \bigcup \mathcal{T}_{\gamma} = \{\varnothing, \{1\}, \{2\}, \{2, 3\}, X\}$. However, we see that while $\{1\}$ and $\{2\}$ are elements of $\mathcal{T}_{\beta} \bigcup \mathcal{T}_{\gamma}$, their union $\{1, 2\}$ is not an element of $\mathcal{T}_{\beta} \bigcup \mathcal{T}_{\gamma}$. This violates the fourth requirement of topologies that unions of open sets be open, so in general the union of topologies is not a topology.

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