General Topology: Closed Sets
Closed Sets
A closed set is a set whose complement is open. Closed sets in topology relate closely to closed sets from the number line from grade school. Recall that such line segments that don't include their endpoints, written $(a,b)$, are "open," while those that do, written $[a,b]$, are "closed." Closed sets in topology provide the scaffolding for this notion of endpoints, which is fleshed out fully in the following section on Closure, Interior, and Boundary.
Problems
Can a set be both open and closed?
Yes. Consider the trivial topology $\mathcal{T} = \{\varnothing, X\}$. Both $X$ and $\varnothing$ are open by definition, and $\varnothing^c = X$ and $X^c = \varnothing$ are both closed.
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Are all closed sets also open?
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Are all open sets also closed?
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Are all sets both open and closed?
Consider the set $X = \{1, 2, 3\}$ and the topology $\mathcal{T}=\{\varnothing, \{1\}, \{2\}, \{1, 2\}, X\}$.
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Are all closed sets also open? No. The set $\{3\}$ is closed because $\{3\}^c=\{1, 2\}$ is open, but $\{3\}$ itself is not open.
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Are all open sets also closed? No. The set $\{1, 2\}$ is open, but its complement $\{3\}$ is not open, therefore $\{1, 2\}$ is itself not closed.
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Are all sets both open and closed? No. The sets in the preceding bullets are counterexamples.
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Show that the intersection of two closed sets is closed.
Let $X_1$ and $X_2$ be closed subsets of $X$. Then $X_1 \cap X_2 = \left(X_1^c \cup X_2^c\right)^c$. By definition of closed set, the sets $X_1^c$ and $X_2^c$ are open. By definition of topology, their union is also open. By definition of closed set again, the complement of their union is closed.
Show that arbitrary intersections of closed sets are closed.
Let $\{A_{\alpha}\}_{\alpha \in I}$ be a collection of closed subsets of $X$. We must show that $\bigcap\limits_{\alpha \in I} A_{\alpha}$ is closed. To do this, we must show that its complement is open. Applying DeMorgan's Law shows us that
$$\left(\bigcap\limits_{\alpha \in I} A_{\alpha}\right)^c = X - \bigcap\limits_{\alpha \in I} A = \bigcup\limits_{\alpha \in I} (X - A_{\alpha}).$$
Each set $X - A_{\alpha}=A_{\alpha}^c$ is open. Arbitrary unions of open sets are open, so the far right side of the equation is open. Therefore its complement $\bigcap\limits_{\alpha \in I} A_{\alpha}$ is closed.
Show that finite unions of closed sets are closed.
Let $A_1, \ldots, A_n$ be closed subsets of $X$. To show that $\bigcup\limits_{i=1}^{n} A_i$ is closed, we must show that its complement is open. Applying DeMorgan's Law shows us that
$$\left(\bigcup\limits_{i=1}^{n} A_i\right)^c = X - \bigcup\limits_{i=1}^{n} A_i = \bigcap\limits_{i=1}^{n}\left(X - A_i\right).$$
Because $A_i$ is closed, $X-A_i=A^c$ is open. Finite unions of open sets are open, so the far right side of the equation is open. Therefore its complement $\bigcup\limits_{i=1}^{n} A_i$ is closed.