2.7 Neighbourhoods, Accumulation Points, Interior Points, Boundary Points
In this section we will discuss neighbourhoods, accumulation points, interior points and boundary
points in topological spaces. As we will see, these are generalizations of the corresponding concepts in
metric spaces.
Definition 2.7.1. Let \(X\) be a topological space and let \(x\) be a point in \(X\). Let \(N\) be a subset of \(X\) which
contains the point \(x\). Then \(N\) is said to be a neighbourhood of \(x\) if and only if there exists an open
set \(U\) such that \(x\in U\) and \(U\subseteq N\).
One can readily verify that this definition of neighbourhoods in
topological spaces is consistent with that of neighbourhoods in metric spaces.
Theorem 2.7.2. Let \(X\) be a topological space. A subset \(U\) of \(X\) is open in \(X\) if and only if \(U\) is a
neighbourhood of each of its points
Proof. Let \(U\) be an open set in \(X\). We must show that \(U\) is a neighbourhood of each of its points.
Let \(x\in U\). We must find an open set \(V\) in \(X\) such that \(x\in V\) and and \(V\subseteq U\). Let \(V = U\) . Then \(x\in V\subseteq U\) and \(V\) is open, so that \(U\) is
a neighbourhood of \(x\). Since \(x\) is an arbitrary element of \(U\), it follows that \(U\) is a neighbourhood of
each of its points.
Now suppose that \(U\) is a neighbourhood of each of its points. Then for every \(x\in U\), there is an open
set \(U_X\) in \(X\) such that \(x\in U_X\subseteq U\). Therefore \(\bigcup _{x\in U}U_X=U\). Since each \(U_X\) is open, it follows that \(U\) is open.
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Example 2.7.3. Given \(X=\{1,2,3\}\) and \(\tau =\{\{1\},\{2\},\{1,2\},X,\emptyset \}\), then \(\tau \) is a topology on \(X\). The set \(\{1,3\}\) is a neighbourhood of 1 but
not a neighbourhood of 3.
We now give the definition of an accumulation point in a topological space.
Definition 2.7.4. Let \(X\) be a topological space and \(A\) a subset of \(X\). A point \(x\in X\) is said to be an
accumulation point of \(A\) if every neighbourhood of \(x\) contains some point of \(A\) other than \(x\).
- a.
- In \(\mathbb {R}\), \(\alpha \) is an accumulation point of \((\alpha ,\beta )\) and also of \([\alpha ,\beta ]\). This illustrates that an accumulation point of a set may or may not be in the set.
- b.
- Let \(A=\{0\}\cup (1,2)\subset \mathbb {R}\). Then 0 is an accumulation point of A (for example B(0,1) does not contain any point of A other than 0). The set of accumulation points in \(\mathbb {R}\) is [1,2]. This illustrates that a point in a set may or may not be an accumulation point of the set.
- c.
- Any point in \(\mathbb {R}\) is an accumulation point of \(\mathbb {Q}\), since for any \(\epsilon >0\), there exists a rational number
in \((x,x+\epsilon )\).
Definition 2.7.6. The closure \(\overline {A}\) of a subset \(A\) of a topological space \(X\) is the union of \(A\) and all the
accumulation points of \(A\) in \(X\).
Example 2.7.7. Let \(X=\{a,b,c\}\). Then \(\tau =\{\{a\},\{b\},\{a,b\},X,\emptyset \}\) is a topology on \(X\). Consider the set \(A=\{a,b\}\). The set of accumulation
points of \(A\) is \(\{c\}\).
Therefore \(\overline {A}=\{a,b,c\}\).
Proposition 2.7.8. A point \(x\) in a topological space \(X\) is in \(\overline {A}\) if and only if every open set containing
\(x\) contains a point of \(A\).
Proof. Let \(x\in \overline {A}\). If \(x\) is an accumulation point of \(A\), then every neighbourhood of \(x\) has a point of \(A\)
other than \(x\). Consequently, every open set containing \(x\) has a point of \(A\) other than \(x\). If \(x\) is not an
accumulation point of \(A\), then \(x\in A\). Therefore if \(U\) is an open set containing \(x\), then \(U\) contains a point of
\(A\).
Conversely, let \(x\in X\), and suppose that every open set containing \(x\) contains a point of \(A\). If \(x\in A\), then \(x\in \overline {A}\). If
\(x\not \in A\), then \(x\) is an accumulation point of \(A\), and hence \(x\in \overline {A}\).
The next theorem gives further properties of the closure of a set in a topological space.
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Theorem 2.7.9. Let \(X\) be a topological space and let \(A\subseteq X\). Then:
- a.
- \(A\) is closed in \(X\) if and only if \(A=\overline {A}\).
- b.
- If \(B\subseteq A\), then \(\overline {B}\subseteq \overline {A}\)
- c.
- \(\overline {\overline {A}}=\overline {A}\)
- d.
- \(\overline {A}\) is closed in \(X\)
Proof.
- a.
- Suppose that \(A\) is closed. Then \(A^c\) is open. If \(x\in X\), then \(x\in A^c\) implies that \(A^c\) is an open set containing
no point of \(A\), so that \(x\not \in \overline {A}\). This implies that \(\overline {A}\subseteq A\). Since \(A\subseteq \overline {A}\), it follows that \(\overline {A}=A\).
Conversely, suppose that \(\overline {A}=A\). Let \(x\in A^c\). Then x is not an accumulation point of A. This implies that there exists an open set \(U_X\) in X such that \(U_X\cap A=\emptyset \) and \(x\in U_X\subseteq A^c\). Hence \(\bigcup _{x\in A^c}U_X=A^c\).
This means that \(A^c\) is open, so that A is closed. - b.
- Suppose that \(x\not \in \overline {A}\). Then \(x\not \in A\) and there is an open set \(U\) in \(X\) such that \(U\cap A=\emptyset \). Since \(B\subseteq A\), it follows that \(x\not \in B\) and \(U\cap B=\emptyset \). This implies that \(x\not \in B\). Hence \(\overline {B}\subseteq \overline {A}\).
- c.
- The inclusion \(\overline {A}\subseteq \overline {\overline {A}}\) is obvious. To show that \(\overline {\overline {A}}\subseteq \overline {A}\), let \(x\in \overline {\overline {A}}\). If \(x\in \overline {A}\), then the inclusion holds.
Suppose that \(\overline {A}\cap U\neq \emptyset \) for every open set \(U\) containing \(x\). Then, since \(A\subseteq \overline {A}\), we have that \(A\cap U\neq \emptyset \). This implies that \(x\in \overline {A}\), consequently \(\overline {\overline {A}}\subseteq \overline {A}\). Hence \(\overline {\overline {A}}=\overline {A}\). - d.
- Let \(x\in (\overline {A})^c\). Then \(x\) is not an accumulation point of \(A\). Thus there exists an open set \(U_X\) such that \(x\in U_X\)
and \(U_X\cap A=\emptyset \). Since \(x\) is not an accumulation point of \(A\), it follows that \(U_X\cap \overline {A}=\emptyset \). From the fact that \(\bigcup _{x\in (\overline {A})^c} U_X=\overline {A}^c\), we get
that \(\overline {A}^c\) is open. Therefore \(\overline {A}\) is closed.
In a topological space \(X\), the closure of a set \(A\) can also be expressed in terms of closed sets in \(X\), as the
following theorem states.
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Theorem 2.7.10. Let \(X\) be a topological space and let \(K\) be the intersection of all the closed sets
containing \(A\) in \(X\). Then \(K=\overline {A}\).
Proof. Since \(A\subseteq \overline {A}\) and \(\overline {A}\) is closed, then \(K\subseteq \overline {A}\). We show that \(\overline {A}\subseteq K\).
Suppose that \(x\not \in K\). Then there is a set F in the collection such that \(x\not \in F\). Since F is closed, \(F^c\) is open.
Now, we have that \(x\in F^c\) and \(F^c\cap F=\emptyset \). Since \(A\subseteq F\), we get that \(F^c\cap A=\emptyset \). This implies that \(x\not \in \overline {A}\). Hence \(\overline {A}\subseteq K\).
In other words, \(\overline {A}\) is the smallest closed set containing \(A\)
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Corollary 2.7.11. Suppose that \(X\) is a topological space and \(A_1,A_2,............,A_n\) are subsets of \(X\). Then \(\overline {A_1\cup A_2\cup .......\cup A_n}=\overline {A_1}\cup \overline {A_2}\cup ...........\cup \overline {A_n}\).
Proof. Let \(A=A_1\cup A_2\cup ...........\cup A_n\). Then \(A_i\subset A\) for all \(i=1,2,............,n\) and so \(\overline {A_i}\subset \overline {A}\) for all \(i=1,2,.........,n\). This implies that
\(\overline {A_1}\cup \overline {A_2}\cup ............\cup \overline {A_n}\subseteq \overline {A_1\cup A_2\cup ...........\cup A_n}\). On the other hand, the set \(\overline {A_1}\cup \overline {A_2}\cup ...........\cup \overline {A_n}\) is closed since it is a finite union of closed sets. Also, \(A=A_1\cup A_2 \cup ..........\cup A_n\subseteq \overline {A_1}\cup \overline {A_2}\cup ..........\cup \overline {A_n}\). It follows
from the previous theorem that \(\overline {A}\subseteq \overline {A_1}\cup \overline {A_2}\cup ...............\cup \overline {A_n}\),
i.e \(\overline {A_1\cup A_2\cup ...........\cup A_n}\subseteq \overline {A_1}\cup \overline {A_2}\cup ...............\cup \overline {A_n}\). Hence the proof.
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Example 2.7.12. In \(\mathbb {R}\). Let \(A_n=[\frac {1}{n},1]\) for \(n\geq 2\). Then \(\overline {A_n}=A_n\) and
\(\bigcup \limits _{n\geq 2}\overline {A_n}=\bigcup \limits _{n\geq 2}A_n=(0,1]\). But \(\Bigg (\overline {\bigcup \limits _{n\geq 2}A_n}\Bigg )=[0,1]\).
This example shows that for arbitrary unions, the previous corollary does not hold.
Definition 2.7.13. Let \(X\) be a topological space. A subset \(A\) of \(X\) is said to be dense in \(X\) if \(\overline {A}=X\).
Example 2.7.14. The set \(\mathbb {Q}\) of rational numbers forms a dense subset of the set of real numbers
\(\mathbb {R}\).
In a topological space, the dual concept of closure in interior, which we define as follows.
Definition 2.7.15. Let \(X\) be a topological space. The interior of a subset \(A\) of \(X\), denoted by \(Int(A)\), is
the union of all subsets of \(A\) which are open in \(X\).
Example 2.7.16. For the set [0,1] in \(\mathbb {R}\), \(Int([0,1])=(0,1)\). In \(\mathbb {R}\). \(Int(\mathbb {Q})=\emptyset \).
Definition 2.7.17. Let \(X\) be a topological space and let \(A\subseteq X\). A point \(x\in X\) is said to be a boundary point
of \(A\) if for every neighbourhood \(N\) of \(x\), \(N\cap A\neq \emptyset \) and \(N\cap A^c\neq \emptyset \). The boundary \(bd(A)\) of the set \(A\) is the set of all boundary
points of \(A\).
Example 2.7.18. The boundary of (0,1] is \(bd((0,1])=\{0,1\}\).
The boundary of \(\mathbb {Q}\) in \(\mathbb {R}\) is \(bd(\mathbb {Q})=\mathbb {R}\).
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