2. Open and Closed Sets

Definition (\(r\)-Neighborhood)

Let \((X,d)\) denote a metric space, \(p\in X\) and \(r>0\). The \(r\)-neighborgood of \(p\) is the set of all \(q\in X\) at distance \(<r\) from \(p\) such that \[ N_r(p):=\left\{q \in X: d(p,q)<r\right\}. \]

• We also have the equivalent term: open ball of radius \(r\) about \(p\), denoted by \(B_r(p)\).

Example

Let \(d(\cdot,\cdot)\) denote a discrete metric defined on the set \(X\), i.e., for any \(x,y\in X\),

1. \(d(x,x)=0\),
2. \(d(x,y)=1\) if \(x\ne y\).

This follows that the \(r\)-neighborhood (open ball of radius \(r>0\)) centered at \(x_0\), \[ N_r(x_0)=\{x\in X:g(x_0,x)<r\} \] can be written as \[ \begin{cases} \{x_0\}\mbox{ }\mbox{ }\mbox{ if }\mbox{ } r\le 1,\\ X \mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ otherwise.} \end{cases} \]

Definition (Interior Point, Open Set)

Let \(E\) be a subset of \(X\). The interior points of \(E\) are those \(p\in X\) such that some neighborhood of which is contained in \(E\), i.e., those \(p\in X\) for which there exists \(r>0\) such that \(N_r(p)\subset E\).

The subset \(E\subset X\) is open if and only if every point of \(E\) is an interior point of \(E\).

Remark

Unlike the notion of boundess, openness of \(E\) depends not only on \(E\), but also on the ambient space \(X\).

For example, every metric space is open as a subset of itself, but a one-point subset of \(\mathbb{R}\) cannot be open as a subset of \(\mathbb{R}\).

Theorem

Every neighborhood is an open set.

• Proof: Define a metric space \((X,d)\). Let \(p\in X\) and let \(r>0\). Then, we must show that for any \(q\in N_r(p)\), there exists an \(h>0\) such that \(N_h(q)\subset N_r(p)\) (every point \(q\in N_r(p)\) is an interior point of \(N_r(p)\)). Define \(h=r-d(p,q)>0\). For any \(s\in N_h(q)\), we have \(s\in N_r(p)\) because \[ d(p,s)\le d(p,q)+d(q,s)<(r-h)+h=r. \] Thus, \(N_h(q)\subset N_r(p)\).

Theorem

1. If \(G_\alpha\) is an open subset of \(X\) for each \(\alpha \in I\), so is \(\cup_{\alpha\in I} G_\alpha\) (countable union).

2. If each of \(G_1,\ldots, G_n\) is an open subset of \(X\), so is \(\cap_{i=1}^n G_i\) (finite intersection).

• Proof of 1: Let \(p\in G:=\cup_{\alpha\in I} G_\alpha\). Then, \(p\in G_\alpha\) for some \(\alpha\in I\). Since \(G_\alpha\) is an open set, \(p \in G_\alpha\) is an interior point, i.e., for every \(p\in G_\alpha\), there exists \(r>0\) s.t. \(N_r(p)\subset G_\alpha\). Since \(G_\alpha\subset G\), \(N_r(p)\subset G\). Because \(p\in G\) is an arbitrary we proved the theorem.

• Proof of 2: Let \(p\in H:=\cap_{i=1}^n G_i\). Then, \(p\in G_i\) for all \(i\). Since \(G_i\) is an open set, \(p\in G_i\) is an interior point, i.e., for every \(p\in G_i\), there exists \(r_i>0\) s.t. \(N_{r_i}(p)\subset G_i\). Let \(r=\min\{r_1,\ldots, r_n\}\). Then, \(r>0\) and \(r\le r_i\) for all \(i\), i.e., which implies \(N_{r}(p)\subset N_{r_i}(p)\) for each \(i\). So, \(N_{r}(p)\subset G_i\) for all \(i\), which implies \(N_{r}(p)\subset H\).

• Counter-example: Someone might wonder why, in the second theorem, we just take into account finite intersection and ignore the case of countable intersection. I will provide a counterexample. Let \(X=\mathbb{R}\) and for all \(n\ge 1\), let \(G_n=(-1/n, 1/n)\). Then, \(\cap_{n=1}^\infty G_{n}=\{0\}\) which is not open.

Definition (Closed Sets, Limit Points)

A closed subset of a metric space \(X\) is defined as the complement of an open subset.

A point \(p\in X\) is said to be a limit point of \(E\subset X\) if every neighborhood of \(p\) contains a point of \(E\) other than \(p\) itself, i.e., every neighborhood of \(p\) contains \(q\in E\), \(q\ne p\), i.e., for every \(r>0\), there exists \(q\in E\) such that \(0<d(p,q)<r\).

• Note that limit points and boundary points are not the same. Consider the set \(S=\{0\}\) in \(\mathbb{R}\) with the usual topology. \(0\) is a boundary point but NOT a limit point of \(S\). Consider the set \(S'=[0,1]\) in \(\mathbb{R}\) with the usual topology. \(0.5\) is a limit point but NOT a boundary point of \(S\).

Theorem

1. If \(G_\alpha\) is a closed subset of \(X\) for each \(\alpha \in I\), so is \(\cap_{\alpha\in I} G_\alpha\) (countable union).

2. If each of \(G_1,\ldots, G_n\) is a closed subset of \(X\), so is \(\cup_{i=1}^n G_i\) (finite intersection).

• Proof : Apply de Morgan’s Law to the previous theorem.

Remark

A subset of a metric space may be both open and closed (clopen) (e.g. in metric space \((X,d)\), \(X\) is open and closed, so \(\phi\) is open and closed). Moreover, there exists a subset which is neither open or closed (e.g. half-open interval \([a,b)\subset \mathbb{R}\), \(\mathbb{Q}\subset \mathbb{R}\)).

Theorem

\(E\subset X\) is closed if and only if every limit point of \(E\) is contained in \(E\).

• (\(\implies\)): Suppose that \(E\) is closed. Let \(x\) be a limit point of \(E\). Assume that \(x\notin E\), so \(x\in E^c\). Since \(E^c\) is open, \(x\) is an interior point of \(E^c\), i.e., there exists \(r>0\) such that \(N_r(x)\subset E^c\). But, because of the definition of the limit point \(x\), every neighborhood of \(x\) should contain at least one point in \(E\), which is contradiction, so \(x\in E\). Because \(x\) is arbitrary, we proved.

• (\(\Longleftarrow\)): Suppose that \(E\) contains all its limit points. Then every \(x\notin E\) is not a limit point of \(E\). So, there exists \(r>0\) such that \(N_r(x)\subset E^c\) contains no point of \(E\), i.e., every point \(x\notin E\) is an interior point of \(E^c\). So, \(E^c\) is open, i.e., \(E\) is closed.

Theorem

A finite set has no limit points.

• Proof: Let \((X,d)\) denote a metric space. For a finite subset \(E\subset X\), let \(p\in E\) denote a limit point of \(E\). Then, by definition, every neighborhood of \(p\) should contain a point \(q\in E\), \(q\ne p\). Because \(E\) is finite, there exist only finite number of \(s \in E\), \(s\ne p\). So, there exist only finite distances \(d(p,s)\). If we choose \(r>0\) smaller than the mininum of the values \(s\ne p\), then the neighborhood of \(p\) with radius \(r>0\) does not contain any point except \(p\).

Definition (Clouse)

Let \(E\) denote a subset of a metric space \(X\). The closure \(\bar E\) of \(E\) is defined to be \(\bar E= E\cup E'\) where \(E'\) is the set of all limit points of \(E\) in \(X\)

Theorem

Let \(A,B\) be subsets of a metric space \(X\)

1. \(A\subset \bar A\),

2. If \(A\subset B\), then \(\bar A\subset \bar B\),

3. \(\bar A\) is closed,

4. The set \(A\) is closed if and only if \(A=\bar A\),

5. The closure of \(\bar A\) is itself, i.e., \(\bar{\bar{A}}= \bar A\).

• Proof of 1: It is clear by the definition of \(\bar A\).

• Proof of 2: Note that \(A\subset B\). Also, \(A'\subset B'\) (\(\because\) let \(p\) denote a limit point in \(A\). Then every neighborhood of \(p\) should contain at least one point \(q \in A\), \(q\ne p\). However, this point \(q\) is also in \(B\) because \(A\subset B\), so \(A'\subset B'\)).

• Proof of 3: Let \(p\in X\), \(p\notin \bar A\). Since \(p\) is not a limit point of \(A\), there exists \(r>0\) such that the neighborhood \(N_r(p)\subset A\) contains no point \(q\in A\) (we get by negation. note that \(q\) is already not equal to \(p\) because \(p\notin \bar A\)). So, \(N_r(p)\) is disjoint from \(A\). We claim that it is also disjoint from \(A'\). Suppose \(q\in N_r(p)\). Since \(N_r(p)\) is open, there exists \(h>0\) s.t. \(N_h(q)\subset N_r(p)\). Thus, \(N_h(q)\) is disjoint from \(A\), and \(q\) is not a limit point of \(A\), as claimed. We conclude that \(N_r(p)\) is disjoint from \(A\cup A'=\bar A\) as desired.

• Proof of 4: (\(\implies\)) Recall the theorem which says “\(A\) is closed if and only if every limit point of \(A\) is contained in \(A\)”. Thus, \(A'\subset A\), so \(\bar A= A\cup A' = A\).

  (\(\Longleftarrow\)) If \(A=\bar A\), then \(A\) is closed by 3.

• Proof of 5: 4 is a direct consequence of 3. Since \(A\) is a closed set by definition, it must be equal to its own closure.

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