Proof. This follows directly from the fact that a sequence in \(\mathbb {R}\) is Cauchy if and only if it converges to a real number. □

Proof. Let \(\{x_k\}^{\infty }_{k=1}\) be Cauchy in \(l^p\), \(x_1=\{x_{11},x_{12},x_{13},\dots \}\), \(x_2=\{x_{21},x_{22},x_{23},\dots \}\),
\(x_3=\{x_{31},x_{32},x_{33},\dots \}\) \(\dots \) and let \(\varepsilon >0\) be given. \[\exists N\in \mathbb {N}\ni k,l>N\implies ||x_k-x_l||<\varepsilon .\] Now, \(\forall n\in \mathbb {N}, |x_{k_{1n}}-x_{l_{1n}}|\leq ||x_k-x_l||_p<\varepsilon .\) Thus \(\{x_{k_{1n}}\}^{\infty }_{k=1}\) is Cauchy in \(\mathbb {R}\) and since \(\mathbb {R}\) is complete, \(\{x_{k_{1n}}\}^{\infty }_{k=1}\) converges to a point \(\alpha _n\in \mathbb {R}\).
Now, let \(\alpha =\{\alpha _n\}^{\infty }_{n=1}\) and let \(\varepsilon >0\) be given. \(\exists N\in \mathbb {N}\ni k,l>N \) \(\implies ||x_k-x_l||_p<\dfrac {\varepsilon }{2}\)
Let \(k>N\), then \(\forall m\in \mathbb {N}\), \begin {align*} \sum ^N_{n=1}|x_{k_{1n}}-x_n|^p &=\sum ^N_{n=1}\Big |x_{k_{1n}}-\lim _{l\rightarrow \infty }x_{l_{1n}}\Big |^p\\ &=\lim _{l\rightarrow \infty }\sum ^N_{n=1}\Big |x_{k_{1n}}-x_{l_{1n}}\Big |^p\\ &\leq \Bigg (\lim _{l\rightarrow \infty }\Bigg (\Bigg (\sum ^{\infty }_{n=1}\Big |x_{k_{1n}}-x_{l_{1n}}\Big |^p\Bigg )^{\frac {1}{p}}\Bigg )^p\\ &=\lim _{l\rightarrow \infty }||x_k-x_l||^p_p\\ &<\Big (\frac {\varepsilon }{2}\Big )^p \end {align*}

Thus, \(\displaystyle {\Bigg (\sum ^N_{n=1}|x_{k_{in}}-\alpha _n|^p\Bigg )^{\frac {1}{p}}<\frac {\varepsilon }{2}}\)

\(\implies ||x_k-\alpha ||_p<\frac {\varepsilon }{2}\) since \(N\) is arbitrary and so \(\alpha =\{\alpha _n\}^{\infty }_{n=1}\in l^p\) and since \(x_n\rightarrow \alpha \in l^p\).

\(\implies l^p\) is complete. □

Proof. Let \(\varepsilon >0\) be given. Then \(\exists N\in \mathbb {N}\ni m,n>N\) \(\implies d(x_n,x_m)<\dfrac {\varepsilon }{2}\hspace {0.2cm}\) and \(d(y_n,y_m)<\dfrac {\varepsilon }{2}.\) Thus \begin {align*} |d(x_n,y_n)-d(x_m,y_m)| &=|d(x_n,y_n)-d(x_n,y_m)+d(x_n,y_m)-d(x_m,y_m)|\\ &\leq |d(x_n,y_n)-d(x_n,y_m)|+|d(x_n,y_m)-d(x_m,y_m)|\\ &\leq d(y_n,y_m)+d(x_n,x_m)\\ &<\frac {\varepsilon }{2}+\frac {\varepsilon }{2}\\ &=\varepsilon \end {align*}

\(\implies \{d(x_n,y_n)\}^{\infty }_{n=1}\) is Cauchy in \(\mathbb {R}\), and since \(\mathbb {R}\) is complete \(\implies d(x_n,y_n)\rightarrow d(x,y)\in \mathbb {R}\). □

Proof. Let \(x\in X\) and \(r>0\).\(\exists x_1\in S_r(x)\) and \(r_1>0\) such that \(r_1\leq \dfrac {1}{2}r\), \(\{y: d(x_1,y)\leq r_1\}\subset S_r(x)\hspace {0.2cm}\) and \(\{y:d(x_1,y)\leq r_1\}\cap A_1=\emptyset \). Likewise, \(\exists x_2\) and \(0<r_2\leq \dfrac {1}{2^2}r\hspace {0.2cm}\) such that \(\{y:d(x_2,y)\leq r_2\}\subset S_{r_1}(x)\) and \(\{y:d(x_2,y)\leq r_2\}\cap (A_1\cup A_2)=\emptyset .\)
Proceeding in the same manner, define sequences \(\{x_n\}^{\infty }_{n=1}\hspace {0.2cm}\) and \(\{r_n\}^{\infty }_{n=1}\hspace {0.2cm}\) such that \(\forall n\in \mathbb {N}\),

i

\(0<r_n<\dfrac {1}{2^n}r\)

ii

\(\{y:d(x_{n+1},y)\leq r_{n+1}\}\subset S_{r_n}(x)\)

iii

\(\{y:d(x_n,y)\leq r_n\}\cap \Big (\bigcup ^n_{i=1}A_i\Big )=\emptyset \)

Then by (i) and (ii), \(\{x_n\}^{\infty }_{n=1}\) is Cauchy sequence in \(X\), and since \(X\) is complete, \(x_n\rightarrow x_0\in X\), since \(x_n\in S_N(x_N) \forall n>N\implies x_0\in S_{r_N}(x_N)\). Now, since \(S_r(x_N)\subset \{y:d(x_N,y)\leq r_N\}\hspace {0.2cm}\) and \(\displaystyle {\{y:d(x_N,y)\leq r_N\}\cap \Big (\bigcup _{i=1}^nA_i\Big )=\emptyset }\) \[\implies x_0\not \in A_N, n\in \mathbb {N}\implies x_0\not \in \bigcup _{n\in \mathbb {N}}A_n\implies x_0\in X/\bigcup _{n\in \mathbb {N}}A_n\] \(X/\bigcup _{n\in \mathbb {N}}A_n\neq \emptyset \)

Since \(x_0\in S_r(x)\implies S_r(x)\cap X/\bigcup _{n\in \mathbb {N}}A_n\neq \emptyset \)

\(\implies x\in \overline {X/\bigcup _{n\in \mathbb {N}}A_n}\)

\(\implies X\subset \overline {X/\bigcup _{n\in \mathbb {N}}A_n}\)

\(\implies X=\overline {X/\bigcup _{n\in \mathbb {N}}A_n}\implies X/\bigcup _{n\in \mathbb {N}}A_n\) is dense in \(X\).

□

Proof. □