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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.4 Closed sets, closure, and interior

Definition 1.4.1. Closed sets.

Let \((X,\mathcal{T})\) be a topological space. A subset \(A\subseteq X\) is closed if its complement \(A^c=X-A\) is open. Equivalently, a subset \(A\) is closed if and only if \(A=U^c\) for some open set \(U\in \mathcal{T}\text{.}\)

Proof.

Statement (1) follows from the observation that \(\emptyset^c=X\text{,}\) and \(X^c=\emptyset\text{.}\)
Assume \(C_i\) is closed for \(1\leq i\leq 2\text{;}\) then by De Morgan’s law
\begin{equation*} (C_1\cup C_2)^c=C_1^c\cap C_2^c\text{,} \end{equation*}
which is open since \(C_i^c\) is open for \(1\leq i\leq 2\) and finite intersections of open sets are open. This proves (2).
Lastly, assume \(C_i\) is closed for all \(i\in I\text{.}\) By De Morgan’s law we have
\begin{equation*} \left(\bigcap_{i\in I}C_i\right)^c=\bigcup_{i\in I}C_i^c\text{,} \end{equation*}
which is open since \(C_i^c\) is open and arbitrary unions of open sets are open.

Remark 1.4.3. Closed sets determine topology.

In general, given a set \(X\) the set complement operation \(Y\mapsto Y^c=X-Y\) is defines a bijection
\begin{align*} \mathcal{P}(X) \amp \leftrightarrow \mathcal{P}(X) \\ Y \amp \mapsto Y^c \text{,} \end{align*}
that is self-inverse: i.e., we have \((Y^c)^c=Y\) for all subsets \(Y\subseteq X\text{.}\) It follows from the Definition 1.4.1 that the closed sets of \(X\) are in bijective correspondence with the open sets, and thus that a topology on \(X\) is uniquely determined by specifying any collection of sets satisfying (1)-(3) from Theorem 1.4.2. In other words, a topology is determined by specifying either its open sets or its closed sets.

Definition 1.4.4. Closed sets axioms.

Let \(X\) be a set. A collection of subsets \(\mathcal{C}\subseteq \mathcal{P}(X)\) is said to satisfy the closed sets axioms if the following properties hold.
  1. Trivial sets.
    \(\emptyset\in \mathcal{C}\) and \(X\in \mathcal{C}\text{.}\)
  2. Finite union.
    If \(C_1, C_2\in \mathcal{C}\text{,}\) then \(\mathcal{C}_1\cup \mathcal{C}_2\in \mathcal{C}\text{.}\)
  3. Arbitrary intersection.
    Given any collection \(\{C_i\}_{i\in I}\) of elements \(C_i\in \mathcal{C}\text{,}\) we have \(C=\bigcap_{i\in I}C_i\in \mathcal{C}\text{.}\)

Proof.

The proof, which is essentially an application of De Morgan’s laws, is left to the reader.

Example 1.4.6. Closed in trivial and discrete topologies.

  1. If \(X\) has the trivial topology, then the only open sets of \(X\) are \(\emptyset, X\text{.}\) Hence the only closed sets are \(\emptyset^c=X\) and \(X^c=\emptyset\text{.}\)
  2. If \(X\) has the discrete topology, then every subset of \(X\) is open. It follows that any subset of \(X\) is closed, since its complement is open. We conclude that al subsets of \(X\) are both open and closed in the discrete topology.

Example 1.4.7. Closed in the cofinite topology.

Let \(X\) be a set endowed with the cofinite topology. An set \(U\) is open in \(X\) if and only if (a) \(U=\emptyset\) or (b) \(U^c\) is finite. It follows that a set \(A\) is closed if and only if (a) \(A=\emptyset^c=X\) or (b) \(A=U^c\) for a cofinite set \(U\) if and only if (a) \(A=X\) or (b) \(A\) is finite.

Proof.

By definition a set \(C\) is closed if and only if its complement \(X-C\) is open. Using (1.2.1), we see that \(X-C\) is open if and only if for all \(x\) there is a basis element \(B\in \mathcal{B} \) satisfying \(x\in B\subseteq X-C\text{.}\)

Example 1.4.9. Closed sets in Euclidean metric topology.

Let \(X=\R^2\) with the Euclidean metric topology. Decide whether the given subset is open and whether it is closed.
  1. \(\displaystyle A=\{(x,y)\colon x\geq 0, y\geq 0\}\)
  2. \(\displaystyle A=\{(x,y)\colon x\geq 0, y > 0\}\)
Solution.
  1. The set \(A\) is not open: any open ball containing the point \((0,1)\) contains points with negative \(x\)-coordinate. Hence there is no open ball containing \((0,1)\) that is contained in \(A\text{.}\)
    The set \(A\) is closed. If \(P=(x,y)\in A^c\text{,}\) then either \(x< 0\) or \(y< 0\text{.}\) Let \(\epsilon=\min\{\abs{x},\abs{y}\}\text{.}\) Every element of \(B_\epsilon(P)\) has either a negative \(x\)-coordinate or a negative \(y\)-coordinate. Hence \(P\in B_{\epsilon}(P)\subseteq A^c\text{.}\) This proves \(A\) is closed, using Lemma 1.4.8.
  2. Again, the set \(A\) is not open, as witnessed by \((0,1)\in A\text{.}\)
    The set \(A^c\) is also not open. Take \(P=(1,0)\in A^c\text{.}\) Any open ball containing \(P\) contains points with positive \(y\)-coordinate. Thus there is no open ball containing \(P\) and contained in \(A^c\text{.}\) We conclude that \(A\) is not closed, using Lemma 1.4.8.

Definition 1.4.10. Interior and closure of a set.

Let \(A\) be a subset of the topological space \(X\text{.}\)
  1. Interior.
    The interior \(A^\circ\) of \(A\) is the union of all open subsets contained in \(A\text{:}\) i.e.,
    \begin{equation*} A^\circ=\bigcup_{U\subseteq A}U, \text{ } U \text{ open}\text{.} \end{equation*}
  2. Closure.
    The closure \(\overline{A}\) is the ntersection of all closed sets containing \(A\text{:}\) i.e.,
    \begin{equation*} \overline{A}=\bigcap_{A\subseteq C}C, \ C \text{ closed}\text{.} \end{equation*}

Proof.

Statements (1)-(2): properties of open (resp. closed) sets imply that \(A^\circ\) is open and \(\overline{A}\) is closed. Furthermore, by definition of \(A^\circ\) (resp. \(\overline{A}\)) along with simple properties of union/intersection it follows that if \(U\) is open and \(U\subseteq A\text{,}\) then \(U\subseteq A^\circ\text{;}\) and if \(C\) is closed containing \(A\text{,}\) then \(\overline{A}\subseteq C\text{.}\)
The first equivalence in (3) follows from the definition of \(A^\circ\) as a union. The second equivalence follows from the fact that any \(U\) is covered by basis elements.
We prove (4) as a chain of equivalences. Call the three statements (i), (ii), (iii).
Assume \(x\in \overline{A}\text{.}\) If \(U\) is an open set satisfying \(U\cap A=\emptyset\text{,}\) then \(C=U^c\) is a closed set containing \(A\text{,}\) and hence contains \(x\text{.}\) It follows that any open set containing \(x\) intersects \(A\) nontrivially. This proves (i) \(\implies\) (ii).
Clearly (ii) implies (iii), since any basis element is an open set.
Assume (iii). If \(C\) is a closed set containing \(A\text{,}\) then \(U=C^c\) is an open set satisfying \(U\cap A=\emptyset\text{.}\) It follows from (iii) that \(x\notin U\text{,}\) and hence that \(x\in C\) for all closed sets \(C\) containing \(A\text{.}\) It follows that \(x\in \overline{A}\text{,}\) the intersection of these sets.

Remark 1.4.12. Using smallest/largest characterization of interior/closure.

Statements (1)-(2) of Theorem 1.4.11 provide a potentially useful indirect way of computing the interior or closure of a set \(A\text{.}\)
  1. Interior.
    Find an open set \(U\) contained in \(A\) and show that for any other open set \(U'\subseteq A\) we have \(U'\subseteq U\text{.}\) It follows that \(U=A^\circ\text{.}\)
  2. Closure.
    Find a closed set \(C\) containing \(A\) and show that for any other closed set \(C'\) containing \(A\) we have \(C'\subseteq C\text{.}\) It follows that \(C=\overline{A}\text{.}\)

Example 1.4.13. Interior and closure of \(K\).

Let \(X=\R\) and let \(K=\{1, 1/2, 1/3,\dots,\}=\{1/N\colon N\in\Z_+\}\text{.}\) Compute the interior and closure of \(K\) with respect to (a) the standard topology on \(\R\) and (b) the \(K\)-topology on \(\R\text{.}\)
Solution.
  1. First consider \(\R\) with the standard topology. The interior of \(K\) is empty in this case: i.e., \(K^\circ=\emptyset\text{.}\) To see this, note that given any \(x=1/N\in K\) and any open interval \((a,b)\) containing \(x\text{,}\) we can choose an \(\epsilon\leq 1/N-1/(N+1)\) such that \((x-\epsilon, x+\epsilon)\subseteq (a,b)\text{;}\) by design all elements of \((x-\epsilon, x+\epsilon)-\{x\}\) lie outside \(K\text{.}\) Thus \(x\notin K^\circ\text{,}\) and \(K^\circ=\emptyset\text{.}\)
    Next. I claim \(\overline{K}=K\cup \{0\}\text{.}\) To see this, note first that \(K\cup \{0\}\) is closed: it is easy to see that for any \(x\notin K\cup\{0\}\) we can find an \(\epsilon\) such that \((x-\epsilon, x+\epsilon)\cap (K\cup \{0\})=\emptyset\text{.}\) By (2) of Theorem 1.4.11 we have \(\overline{K}\subseteq K\cup \{0\}\text{.}\) Since furthermore \(K\subseteq \overline{K}\subseteq K\cup \{0\}\) and since \(K\) itself is not closed, it follows that \(\overline{K}=K\cup \{0\}\text{.}\)
  2. Now consider the \(K\)-topology on \(\R\text{.}\) We still have \(K^\circ=\emptyset\text{.}\) Indeed, fix any \(x=1/N\in K\text{,}\) and take any basis element \(B\) containing \(x\text{.}\) We have \(B=(a,b)\) or \(B=(a,b)-K\) for some open interval \((a,b)\text{.}\) In either case, the same argument as above shows that \(B\) contains a basis element \(B'=(x-\epsilon, x+\epsilon)-\{x\}\text{,}\) all of whose elements lie outside of \(K\text{.}\) This shows \(x\notin K^\circ\text{,}\) and hence \(K^\circ=\emptyset\text{.}\)
    Next, I claim that \(K\) is closed in the \(K\)-topology, from whence it follows that \(K=\overline{K}\text{.}\) (Use (2) from Theorem 1.4.11.) To prove the claim, we must show, using Lemma 1.4.8 that for any \(x\notin K\) there is a basis element (in the \(K\)-topology) containing \(x\) and contained in \(K^c\text{.}\) There are two cases: \(x\ne 0\) and \(x=0\text{.}\) If \(x\ne 0\text{,}\) we can find an open interval \((a,b)\) such that \(x\in (a,b)\) and \((a,b)\cap K=\emptyset\text{.}\) Since open intervals are basis elements in the \(K\)-topology, we are done in this case. Next, assume \(x=0\text{.}\) In this case the basis element \(B=(-1,1)-K\) satisfies \(x\in B\subseteq K^c\text{.}\) Having exhausted the cases, we conclude \(K\) is closed, and hence \(\overline{K}=K\text{.}\)

Example 1.4.14. Interior and closure in cofinite topology.

Let \(X\) be a set endowed with the cofinite topology. For \(A\subseteq X\) describe \(A^\circ\) and \(\overline{A}\text{.}\) Use cases.
Solution.
If \(X\) is finite, then the cofinite topology is equal to the discrete topology. In this case \(\overline{A}=A^\circ=A\) for any subset \(A\subseteq X\text{.}\)
Now assume \(X\) is infinite. If \(A\in \{\emptyset, X\}\text{,}\) then \(\overline{A}=A^\circ=A\text{,}\) since such an \(A\) is both open and closed. Assume \(\emptyset\subsetneq A\subsetneq X\text{.}\) Recall that nonempty open sets of \(X\) are precisely the cofinite ones, and that the closed sets that precisely the finite ones (or else \(X\) itself). Thus we have
\begin{align*} \overline{A}\amp=\bigcap_{\substack{C \text{ closed}\\ A\subseteq C}}C=X\cap \bigcap_{\substack{C \text{ finite}\\ A\subseteq C}}C=\begin{cases} X \amp \text{if } A \text{ infinite} \\ A \amp \text{if } A \text{ finite}. \end{cases} \\ A^\circ \amp =\bigcup_{\substack{U \text{ open}\\ U\subseteq A}}U =\emptyset \cup \bigcup_{\substack{U \text{ cofinite}\\ U\subseteq A}}U =\begin{cases} \emptyset \amp \text{if } A \text{ not cofinite}\\ A \amp \text{if } A \text{ cofinite} \end{cases}\text{.} \end{align*}
Note, for the equalities above we use the fact that if \(A\) is infinite (i.e., not finite), then it is not contained in any finite set; and if \(A\) is not cofinite, then it does not contain any cofinite set.
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