Proof. If we take \(x_0=\dfrac {a+b}{2}\) and \(r=\dfrac {|a-b|}{2}\), then the open ball interval (a,b) is the open ball \(B(x_0,r)\) and the closed interval \([a,b]\) is the closed ball \(B[x_0,r]\).
Furthermore, if a and b are distinct real numbers, we obtain the sphere \(S(x_0,r)=\{x\in \mathbb {R}:|x-y|=r\}=\{a,b\}\).


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Solution.

i

\begin {align*} B((0,0),1) &=\{(x_1,x_2)\in \mathbb {R}^2:d_1((0,0),(x_1,x_2))<1\}\\ &=\{(x_1,x_2)\in \mathbb {R}^2:|x_1|+|x_2|<1\} \end {align*}

This set represents the interior of the region bounded by the lines \(x_1+x_2=1\), \(x_1-x_2=1\), \(-x_1+x_2=1\) and \(-x_1-x_2=1\)

ii

\begin {align*} B((0,0),1) &=\{(x_1,x_2)\in \mathbb {R}^2:d_2((x_1,x_2),(0,0))<1\}\\ &=\{(x_1,x_2)\in \mathbb {R}^2:\sqrt {x_1^2+x^2_2}<1\} \end {align*}

This represents the interior of the circle of radius 1 centered at the origin.

iii

\begin {align*} B((0,0),1) &=\{(x_1,x_2)\in \mathbb {R}^2:d_3((x_1,x_2),(0,0))<1\}\\ &=\{(x_1,x_2)\in \mathbb {R}^2:\max \{|x_1|,|x_2|<1\}\} \end {align*}

This represents the interior of the square of diameter 2 centered at the origin.

Proof. Let \(x\in A\), where A is a union of open intervals in \(\mathbb {R}\). Then X belongs to atleast one open intervals. Since every open interval in \(\mathbb {R}\) is an open ball, it follows that A is an open set.

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Proof. Let \(\{O_{\lambda }:\lambda \in \Omega \}\) be any family of open sets in X and let \(x\in O=\bigcup _{\lambda \in \Omega } O_{\lambda }\). Then \(x\in O_{\lambda }\), for at least one \(\lambda \in \Omega \). Since \(O_{\lambda }\) is open in X, there is an open ball B(x,r) such that \(B(x,r)\subseteq O_{\lambda }\). Since \(O_{\lambda } \subseteq O\), it follows that \(B(x,r)\subseteq O\). Since x was an arbitrary element of O, we obtain that O is an open set in X.
To prove the second part, suppose that \(O_1,O_2,...........,O_n\) is a finite
collection of open sets in X and \(O=\bigcap _{k=1}^{n} O_k\). If \(x\in O\), then \(x\in O_k\) for all \(k=1,2,...........n\). Since each \(O_k\) is an open set in X, there is an open ball \(B(x,r_k)\) such that \(B(x,r_k)\subseteq O_k\) for all \(k=1,2,.........,n\). Let
\(r=\min \{r_1, r_2,............,r_n\}\). Then \(B(x,r)\subseteq B(x,r_k)\) for all \(k=1,2,.........,n\). It follows that \(B(x,r)\subseteq O_k\) for all \(k=1,2,..........,n\), so that \(B(x,r)\subseteq O\). Hence O is open in X.

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Proof. To show that \(\emptyset \) is an open set in X, we need to show that every point in \(\emptyset \) is contained in an open ball which is a subset of \(\emptyset \). But this holds vacuously since \(\emptyset \) has no elements.
To prove the second part, if \(x\in X\), then every open ball \(B(x,r)\) is a subset of X. Hence X is an open set.

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Proof. Suppose that \(O\) is an open set in \(X\) and \(x_0\in O\). Then there is a real number \(r>0\) such that \(B(x_0,r)\subseteq O\), so that \(O\) is a neighbourhood of \(x_0\). Since \(x_0\) was an arbitrary point in \(O\), we have that \(O\) is a neighbourhood of each of its points.
Conversely, suppose that \(O\) is a neighbourhood of each of its points. If \(x_0\in O\), then there is a real number \(r>0\) such that \(B(x_0,r)\subseteq O\). Since \(x_0\) was an arbitrary point in \(O\), it follows that \(O\) is open in X.

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Proof. Let \(\{N_{\lambda }:\lambda \in \Omega \}\) be a collection of neighbourhoods of \(x_0\) and let
\(N=\bigcup _{\lambda \in \Omega } N_{\lambda }\). Since \(N_{\lambda }\) is a neighbourhood of \(x_0\) for each \(\lambda \in \Omega \), then for each \(\lambda \), there exists an open ball \(B(x_0,r)\) in X such that \[x_0\in B(x_0,r_{\lambda })\subseteq N_{\lambda }\subseteq \bigcup _{\lambda \in \Omega } N_{\lambda }=N\] This means that \(N\) is a neighbourhood of \(x_0\).
To prove the second part, suppose that \(N_1,N_2,.......,N_n\) is a finite collection of neighbourhoods of \(x_0\) and \(N=\bigcap ^{n}_{k=1} N_k\). Then there exists open balls \(B(x_0,r_1), B(x_0,r_2),.......,B(x_0,r_n)\) such that \(B(x_0,r_k)\subseteq N_k\) for \(k=1,2,......,n\). Let \(r=\min \{r_1, r_2,.....,r_n\}\). Then \(B(x_0,r)\subseteq B(x_0,r_k)\) for \(k=1,2,.....,n\), so that \(x_0\in B(x_0,r)\subseteq \bigcap ^n_{k=1} N_k =N\). Hence N is a neighbourhood of \(x_0\).

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Proof. Let \(a\in [0,1)\). If \(a=0\), then there is no \(r>0\) such that \(B(a,r)\subseteq [0,1)\). Thus \(a\not \in Int(A)\). Suppose that \(a\neq 0\). Take \(r=\min \{\frac {a}{2},\frac {1-a}{2}\}\). Then \(B(a,r)\subseteq [0,1)\) and so \(a\in Int(A)\). Hence \(Int(A)=(0,1)\).


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Proof. We prove that every open set contained in \(A\) is contained in \(Int(A)\). To this end suppose that \(O\) is an open set contained in \(A\). Then \(O\) is a neighbourhood of each of its points and since \(O\subseteq A\), we have that \(A\) is a neighbourhood of each of the points of \(O\). This implies that \(O\subseteq Int(A)\).

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Solution. If \(r>0\) is any real number, \(B(0,r)\) contains points in both \(A\) and \(A^c\). Therefore \(0\in bd(A)\). Similarly, \(1\in bd(A)\). Now consider \(x\in A\) such that \(x\neq 0\) and \(x\neq 1\). Take \(r=\min \{\frac {x}{2},\frac {1-x}{2}\}\). Then \(B(x,r)\subseteq A\) and so x is not a boundary point of A. This means that \(bd(A)=\{0,1\}\).

Note in the previous example that a point in the set \(A\) is either a boundary point or an interior point of \(A\). The next result shows that this is true for every subset of a metric space.

Proof. Let \(x\in A\). If there is a real number \(r>0\) such that \(B(x,r)\subseteq A\), then \(x \)is an interior point of \(A\) and the result holds. If there is no real number \(r>0\) such that \(B(x,r)\subseteq A\), then \(x\) is a boundary point of \(A\) and in this case the result also holds.


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