1.3 Accumulation Points, Closed Sets, Closure
We have so far introduced some basic concepts associated with metric spaces. In this section, we
develop further concepts on metric spaces.
Definition 1.3.1. Let \((X,d)\) be a metric space and let \(A\) be a subset of \(X\). A point \(x_0\) of \(X\) is said to be an
accumulation point of \(A\) if every open ball \(B(x_0,r)\) contains a point of \(A\) other than \(x_0\). A point \(y_0\) in \(A\) is said
to be an isolated point of \(A\) if \(y_0\) is not an accumulation point of \(A\). Accumulation points of set A in
metric space are sometimes called limit points.
Note that an accumulation point of a subset of a metric space may not belong to the set.
Example 1.3.2. Let \(A=\{-1\}\cup (0,1)\) be a subset of the metric space \((\mathbb {R},d)\) where d is the usual metric.
Find the accumulation points and isolated points of \(A\).
Solution. For any real number \(r\leq 1\), the open ball \(B(-1,r)\) does not contain any point of \(A\). Thus -1 is an
isolated point of \(A\). On the other hand if \(x\in [0,1]\), then every open ball \(B(x,r)\) contains a point of A. Therefore
every point in [0,1] is an accumulation point of \(A\). Hence, the set of isolated points of is \(\{-1\}\) and the
set of accumulation points is [0,1].
Definition 1.3.3. A subset \(F\) of a metric space \((X,d)\) is said to be closed if \(F\) contains all its
accumulation points.
Example 1.3.4. The set \(F=[0,1]\) is closed in the metric space \((\mathbb {R},d)\) where d is the usual metric on \(\mathbb {R}\).
Proof. The set of accumulation points of \(F\) is itself. Hence contains all its accumulation points
and is therefore closed.
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Definition 1.3.5. Let \(A\) be a subset of a metric space \((X,d)\). The union of \(A\) and the set of all the
accumulation points of \(A\) is called the closure of \(A\).
The closure of \(A\) is denoted by \(\overline {A}\). From the definition it is clear that \(A\subseteq \overline {A}\).
Theorem 1.3.6. A subset \(F\) of a metric space (X,d) is closed if and only if its complement is
open.
Proof. Suppose that \(F\) is closed in \(X\), let \(x\in F^c\). Since \(F\) is closed, it contains all its accumulation points
and so \(x\) is not an accumulation point of \(F\). This implies that there exists a real number \(r>0\) such that
\(B(x,r)\subseteq F^c\). Since \(x\) was an arbitrary element of \(F^c\), it follows that \(F^c\) is open.
Conversely, suppose that \(F^c\) is open. Let \(x\) be an accumulation point of \(F\). We show that \(x\in F\). If \(x\in F^c\), then
since \(F^c\) is open, there exist \(r>0\) such that \(B(x,r)\subseteq F^c\). Thus this means that \(B(x,r)\cap F=\emptyset \). Which contradicts the fact that \(x\)
is an accumulation point of \(F\). Hence \(x\in F\) and so \(F\) is closed.
The next theorem shows that every closed ball in a metric space is a closed set.
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Proof. We show that \((B[x,r])^c\) is open in X. If \((B[x,r])^c\) is empty, then \((B[x,r])^c\) is open and we are done. Assume that \((B[x,r])^c\) is
non-empty. Let
\(y\in (B[x,r])^c\). Then \(d(x,y)>r\) and so \(r_1=d(x,y)-r>0\). We show that \(B(y,r_1)\subseteq (B[x,r])^c\). If \(z\in B(y,r_1)\) then \(d(y,z)<r_1\). It follows from the triangle inequality that \(d(x,y)<d(x,z)+d(z,y)\), and so
\begin {align*} d(x,z)&>d(x,y)-d(z,y)\\ &>d(x,y)-r_1\\ &=d(x,y)-[d(x,y)-r]\\ &=r \end {align*}
Thus \(d(x,z)>r\) and so \(z\in (B[x,r])^c\). Hence \((B[x,r])^c\) is open from which we obtain that \(B[x,r]\) is closed.
The next theorem provides some of the properties of closed sets.
□
- i
- \(\emptyset \) and \(X\) are closed sets
- ii
- the union of any finite collection of closed sets is closed.
- iii
- the intersection of any collection of closed sets is closed.
Proof.
- i
- The set \(\emptyset \) has no points and so it has no accumulation points, therefore \(\emptyset \) vacuously contains
all its accumulation points and it is closed.
On the other hand, since X is the whole space it contains all its
accumulation points and therefore closed. - ii
- Let \(F_1,F_2,..........,F_n\) be any finite collection of closed sets and let
\(F=\bigcup ^{n}_{k=1} F_k\). We show that \(F^c\) is open. Since \(F_k\) is closed for all
\(k=1,2,........,n\); \(F^c_k\) is open for all \(k=1,2,...........,n\). Since
\(F^c=\bigcap ^{n}_{k=1} F_k^c\) by De-Morgan’s theorem, and since the intersection of a finite number of open sets is open, it follows that \(F^c\) is open. Hence F is closed. - iii
- Let \(\{F_{\lambda }:\lambda \in \Omega \}\) be any collection of closed sets in X and let
\(F=\bigcap _{\lambda \in \Omega } F_{\lambda }\). We need to show that \(F^c\) is open. By De-Morgan’s theorem, we have that \(F^c=\bigcup _{\lambda \in \Omega } F^c_{\lambda }\). Since \(F_{\lambda }\) is closed for all \(\lambda \in \Omega \), \(F^c_{\lambda }\) is open for all \(\lambda \in \Omega \). Since the union of any collection of open sets is open \(F^c\) is open, and so F is closed.
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