Technology and Art

Avishek Sen Gupta on 25 October 2021

This post lists solutions to the exercises in the **Completion of Metric Spaces section 1.6** of *Erwin Kreyszig’s* **Introductory Functional Analysis with Applications**. This is a work in progress, and proofs may be refined over time.

**Proof:**

By definition, a limit point \(L\) of the set \(Y\) has at least one point \(x \neq L\) within every neightbourhood \(\epsilon\). Since \(Y\) has a finite number of points, then it has no limit points, and thus (vacuously) contains all its limit points.

Thus, \(Y\) is a complete metric subspace.

\[\blacksquare\]**Proof:**
The completion of \((X,d)\), where \(X\) is the set of all rational numbers and \(d(x,y)=\vert x-y \vert\), is \(\mathbb{R}\), since every real number is the limit of a sequence of rational numbers.

**Proof:**
A discrete metric space \(X\) has no limit points, since no point in it has at least one point in every neighbourhood \(\epsilon\). Thus, it vacuously contains all its limit points. Thus, the completion of the discrete metric space \(X\) is itself.

**Proof:**

Assume \(x,y \in X_1\), and let \(T:X_1 \rightarrow X_2\). Since \(X_1\) and \(X_2\) are isometric, we have:

\[d(x,y)=\bar{d}(Tx,Ty)\]We know that \(X_1\) is complete: let us assume, for an arbitrary \(\epsilon\), a point \(x_1 \in X_1\) lying in the \(\epsilon\)-neighbourhood of a limit point \(x \in X_1\). Then, we have:

\[d(x,x_1)=d(Tx,Tx_1) < \epsilon\]Thus, for an arbitrary \(\epsilon\), there is a point \(Tx \in X_2\) which has a point \(Tx_1\) in its \(\epsilon\)-neighbourhood as well. Thus, \(Tx\) is a limit point of \(X_2\) as well. Since \(Tx \in X_2\) for all \(x \in X_1\), \(X_2\) contains all its limit points as well.

Thus, \(X_2\) is complete.

\[\blacksquare\]**Proof:**

Consider a Cauchy sequence \((x_n)\) in \(X\). Then, we have, \(\forall \epsilon>0\), \(\exists N\), such that \(d(x_m,x_n) < \epsilon\) for all \(m,n>N\).

Let \(T:X \rightarrow Y\). Since \(X\) is isometric to \(Y\), we have:

\[d(x_m,x_n)=d(Tx_m,Tx_n) < \epsilon\]This implies that for every \(\epsilon>0\), there exists a \(\delta>0\), such that \(d(x_m,x_n) < \delta \Rightarrow d(Tx_m,Tx_n) < \epsilon\). In this case \(\delta=\epsilon\). Thus, \(T\) is continuous at \(x_n\).

The above argument can be used for \(T^{-1}\) to prove that it is also continuous.

To prove injectivity, we note that \(x \neq y \Rightarrow d(x,y) \neq 0 \Rightarrow \Rightarrow d(Tx,Ty) \neq 0 \Rightarrow Tx \neq Ty\).

To prove surjectivity, we pick a point \(y \in Y\). Assume \(x_1 \in X\). Then, by isometry we must have: d(y,Tx_1)=d(x, x_1), where \(x \in X\). Thus, there is a corresponding preimage for every \(y \in Y\).

\[\blacksquare\]Consider the \(f:(0,1) \rightarrow \mathbb{R}\) defined as \(f(x)=x\). Then \(f(x)\) and its inverse are continuous and bijective. \((0,1)\) is an incomplete metric space and \(\mathbb{R}\) is complete.

**Proof:**

We note that \(f(t)=\displaystyle\frac{t-a}{b-a}, a \neq b\) is a mapping \(f: [a,b] \rightarrow [0,1]\), and that \(f^{-1}(t)=a+(b-a)t\) is a mapping \(f^{-1}: [0,1] \rightarrow [a,b]\).

We note that \(f\) and \(f^{-1}\) are bijections. The distance metric in \(C\) is defined as \(d(x,y)=\sup\vert x(t) - y(t) \vert\).

Define a mapping \(T:C_{t \in [0,1]}(t) \rightarrow C_{t \in [a,b]}(f^{-1}(t))\)

Think of \(C(f(t)\) as the original function applied to \([0,1]\) even though the input \(t \in [a,b]\). Then, practically we have \(C_{t \in [0,1]}(t)=C_{t \in [a,b]}(f(t))\).

Then:

\[d(Tx,Ty)=\sup_{[a,b]} |x(f(t)) - y(f(t))|=\sup_{[0,1]} |x(t) - y(t)|=d(x,y)\]Thus, \(T\) preserves distances.

To prove injectivity, suppose \(Tx=Ty\), then we have:

\[d(Tx,Ty)=\sup_{[a,b]} |x(f(t)) - y(f(t))|=0 \\ \Rightarrow \sup_{[0,1]} |x(t) - y(t)|=d(x,y)=0 \\ \Rightarrow x=y\]For surjectivity, we note that for an arbitrary function \(y(f(t)) \in C[a,b]\), we always have \(x(t) \in C[0,1]\), since \(T^{-1}x=x(f^{-1}(f(t)))=x(t)\).

\[\blacksquare\]**Proof:**

Let \((x_n)\) be a Cauchy sequence in \((X,\bar{d})\), so that we have, \(\forall \epsilon>0, \exists N\), such that \(d(x_m,x_n) < \epsilon\) for all \(m,n>N\).

Then, we have:

\[\frac{d(x_m,x_n)}{1+d(x_m,x_n)} < \epsilon \\ d(x_m,x_n) < \epsilon + \epsilon d(x_m,x_n) \\ d(x_m,x_n)(1-\epsilon) < \epsilon \\ d(x_m,x_n) < \frac{\epsilon}{1-\epsilon}\]Set \(\epsilon=\frac{1}{k}\), so that we get:

\[{d}(x_m,x_n) < \frac{1}{k-1} \\\]\(k\) can be made as large as needed to make \(\epsilon\) as small as needed. Thus, the sequence \((x_n)\) is Cauchy in \((X,d)\), and thus has a limit \(x\), i.e., \(x_n \rightarrow x\).

Then, \(d(x_n,x)<\epsilon\).

\[\bar{d}(x_n,x)<d(x_n,x)<\epsilon\] \[\blacksquare\]**Proof:**
Suppose \((X,\tilde{d})\) is complete. Then, we have:

**Proof:**

(1) defines equivalence of two sequences as \((x_n)\tilde(x_n') \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0\).

**Proof:**

Since \((x_n)\) and \((x_n')\) are convergent, we have, \(\forall \epsilon>0, \exists N_1, N_2\) such that \(d(x_m,l)<\epsilon\) and \(d(x_n',l)<\epsilon\), for \(m>N_1, n>N_2\). Choose \(N=\max(N_1,N_2)\), so that we have \(d(x_n,l)<\epsilon\) and \(d(x_n',l)<\epsilon\) for all \(n>N\).

\[d(x_n,x_n') \leq d(x_n,l) + d(l,x_n') < \epsilon+\epsilon=2 \epsilon \\ \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n') = 0\] \[\blacksquare\](1) defines equivalence of two sequences as \((x_n)\tilde{}(x_n') \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0\).

**Proof:**

We will check for the following properties:

- Reflexive
- Symmetric
- Transitive

We know that \(d(x_n,x_n)=0\) always because of the **Principle of Indiscernibles**. Thus, we get:

By the **Symmetry Property** of a distance metric, we know that \(d(x_n,x_n')=d(x_n',x_n)\). Thus if we have \(\lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0\), then we also have:

By the **Triangle Inequality**, we have:

Taking limits, we get:

\[\lim\limits_{n \rightarrow \infty} d(x_n,z_n) \leq \lim\limits_{n \rightarrow \infty} d(x_n,y_n) + \lim\limits_{n \rightarrow \infty} d(y_n,z_n)\]If we have \(\lim\limits_{n \rightarrow \infty} d(x_n,y_n)=0\) and \(\lim\limits_{n \rightarrow \infty} d(y_n,z_n)=0\), we get:

\[\lim\limits_{n \rightarrow \infty} d(x_n,z_n) \leq 0\]Since distances are always nonnegative, we have: \(\lim\limits_{n \rightarrow \infty} d(x_n,z_n) = 0\).

\[\blacksquare\]**Proof:**

Since \((x_n)\) is Cauchy, we have, \(\forall \epsilon>0, \exists N\) such that \(d(x_m,x_n)<\epsilon\) for \(m,n>N\).

\[d(x_m,x_n) < \epsilon\]We also have \((x_n)\) and \((x_n')\) being equivalent, so we can write:

\[\lim\limits_{n \rightarrow \infty} d(x_n,x_n') = 0\]By the **Triangle Inequality**, we have:

Taking limits on both sides, we get:

\[\lim\limits_{n \rightarrow \infty} d(x_m',x_n') \leq \underbrace{\lim\limits_{n \rightarrow \infty} d(x_m',x_m)}_\text{0 because equivalent} + \underbrace{\lim\limits_{n \rightarrow \infty} d(x_m,x_n)}_\text{0 because Cauchy} + \underbrace{\lim\limits_{n \rightarrow \infty} d(x_n,x_n')}_\text{0 because equivalent} \\\]Since distance metric has to be nonnegative, we conclude that:

\[\lim\limits_{n \rightarrow \infty} d(x_m',x_n')=0\] \[\blacksquare\]**Proof:**

**(M1)** We know that \(d(x,y)=\vert \xi_1 - \eta_1\vert\) is always nonnegative, real-valued, and finite.

**(M3)** Because \(d(x,y)=\vert \xi_1 - \eta_1\vert\) has a modulus sign, we always have: \(\vert \xi_1 - \eta_1\vert = \vert \eta_1 - \xi_1\vert\), and thus we have symmetry.

**(M4)** We have: \(d(x,y)=\vert \xi_1 - \eta_1\vert = \vert \xi_1 - \kappa_1 + \kappa_1 - \eta_1\vert \leq \vert \xi_1 - \kappa_1 \vert + \vert \kappa_1 - \eta_1\vert\). Thus, the **Triangle Inequaity** is shown.

**(Modified M2)** An example pair which satisfies this condition is \((1,2)\) and \((1,3)\). We see that any pair \((\kappa,\xi)\) and \((\kappa,\eta)\) will satisfy **(Modified M2)**. We see that if \(x_1=(\kappa,\xi)\) and \(x_2=(\kappa,\eta)\), then \(d(x,y)=\vert \kappa - \kappa\vert=0\).

**Proof:**

**[TODO]**

**Answer:**

The open ball in this case is a vertical rectangles with open width 2 centered at \(x_0\).

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