Technology and Art

Avishek Sen Gupta on 12 October 2021

This post lists solutions to the exercises in the **Convergence, Cauchy Sequences, and Completeness section 1.4** of *Erwin Kreyszig’s* **Introductory Functional Analysis with Applications**. This is a work in progress, and proofs may be refined over time.

**Proof:**

Suppose \((x_n)\) is convergent and \(x_n \rightarrow x\). Let \((x_{n_k})\) be a subsequence. Let \(x_{n_m}\) correspond to \(x_i\).

Since \((x_n)\) is convergent, \(\forall \epsilon>0, \exists N\), such that \(d(x_n,x)<\epsilon\) for \(n>N\). Pick \(p \geq N\), such that \(x_p\) exists in \((x_{n_k})\) and is identified as \(x_{n_m}\).

This implies that \(\forall \epsilon>0, \exists M\), such that \(d(x_{n_m},x)<\epsilon\).

The above is the definition of the convergence of a sequence to a limit. Thus, \((x_{n_k})\) converges to \(x\).

Alternatively, you can prove that the limit of \((x_{n_k})\) is \(x\) in the following manner.

Suppose \(x_{n_k} \rightarrow y\). Then, by the **Triangle Inequality**, we have:

Both \(d(x,x_p)\) and \(d(x_{n_m},y)\) can be made as small as possible, since \((x_n)\) and \((x_{n_k})\) converge, implying that \(d(x,y)\) is smaller than any positive value. Thus:

\[d(x,y) \leq d(x,x_p) + d(x_{n_m},y) \\ \Rightarrow d(x,y) < \epsilon_1 + \epsilon_2, \epsilon_1, \epsilon_2 > 0 \\ \Rightarrow d(x,y)=0\]Thus, \(x=y\), i.e., \(x_{n_k}\) has the same limit as \((x_n)\).

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

Suppose \((x_n)\) is Cauchy. Let \((x_{n_k})\) be a subsequence. Let \(x_{n_m}\) correspond to \(x_i\).

- Since \((x_{n_k})\) is convergent, \(\forall \epsilon>0, \exists M\), such that \(d(x_{n_k},x)<\epsilon\) for \(n>M\).
- Since \((x_n)\) is convergent, \(\forall \epsilon>0, \exists N\), such that \(d(x_i,x_j)<\epsilon\) for \(i,j>N\).

Pick \(N_0=\max(M,N)\). Then the above two statements become:

- Since \((x_{n_k})\) is convergent, \(\forall \epsilon>0, \exists N_0\), such that \(d(x_{n_i},x)<\epsilon\) for \(i>N_0\).
- Since \((x_n)\) is convergent, \(\forall \epsilon>0, \exists N_0\), such that \(d(x_i,x_j)<\epsilon\) for \(i,j>N_0\).

(Note that we picked the same \(i\) for both \((x_n)\) and \((x_{n_k})\) because any value greater than \(N_0\) will fulfil the above conditions, so we might as well pick the same index. For any index \(i\), we can use \(x_i\) and \(x_{n_i}\) interchangeably, since they index the same element in both the sequence and the subsequence.)

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

\(\epsilon\) can be made as small as possible, implying that \(d(x_j,x)\) is smaller than any positive value. Thus:

\[d(x_j,x)=0\]Hence, \((x_n)\) is convergent with the limit \(x\).

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

Suppose that \(x_n \rightarrow x\). Then \(\forall \epsilon>0, \exists N_0\), such that \(d(x_n,x)<\epsilon\) for all \(n>N_0\). This implies that a neighbourhood \(V_\epsilon\) of \(x\) exists, which contains all \(x_{n>N_0}\). Since there are an infinite number of values for \(\epsilon\), it follows that this applies to every neighbourhood of \(x\).

Conversely, suppose that for every neighborhood \(V\) of \(x\) there is an integer \(n_0\) such that \(x_n \in V\) for all \(n > n_0\). Assume each neighbourhood has a size of \(\epsilon\). Thus, \(x_n \in V\) implies that \(d(x_n,x)<\epsilon\). Then, we can restate this as the following: \(\forall \epsilon>0, \exists N_0\), such that \(d(x_n,x)<\epsilon\) for all \(n>N_0\).

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

By definition, for a Cauchy sequence, we have: \(\forall \epsilon>0, \exists N_0\), such that \(d(x_m,x_n)<\epsilon\) for all $m,n>N_0$$.

Choose \(\epsilon=1\). Then, assume the value of \(N_0\) to be \(N_1\). For any \(d(x_a,x_b)\), we have:

- \(a<b \leq N_0\): Then \(d(x_a,x_b) \leq a = max[d(x_a,x_0), d(x_a,x_1), \cdots, d(x_a,x_{N_1})]\)
- \(N_0<a<b\): Then \(d(x_a,x_b) < \epsilon = 1\)
- \(a \leq N_0<b\): By the
**Triangle Inequality**, we have: \(d(x_a,x_b) \leq d(x_a,x_{N_1}) + d(x_{N_1}, x_b) < a + 1\)

Combining these upper bounds, we get: \(\sup d(x_a,x_b) < a+1\)

\[\blacksquare\]**Answer:**

Consider the discrete metric on \(\mathbb{R}\). If we have a sequence \((x_n)=0,1,0,1,\cdots\), then the series is bounded because \(\sup d(x_m, x_n)=1\), but for \(\epsilon=\frac{1}{2}\), there is no \(N\) for which \(d(x_m,x_n)<\epsilon\) for \(m,n>N\). Thus, the sequence is not Cauchy, though it is bounded.

Convergence is sufficient for a sequence to be Cauchy. For convergence, we have the condition: if \(x_n \rightarrow x\), \(\forall \epsilon>0, \exists N_0\), such that \(d(x_n,x)<\epsilon\) for all \(n>N_0\).

Consider \(m,n>N_0\). Then, by the **Triangle Inequality**, we have:

thus proving the Cauchy criterion.

**Proof:**

Then, we have:

\[d(x_m,y_m) \leq d(x_m,x_n) + d(x_n,y_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) \leq d(x_m,x_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) < 2 \epsilon\]Similarly, we have:

\[d(x_n,y_n) \leq d(x_n,x_m) + d(x_m,y_m) + d(y_m,y_n) \\ d(x_n,y_n) - d(x_m,y_m) \leq d(x_n,x_m) + d(y_m,y_n) \\ \Rightarrow d(x_n,y_n) - d(x_m,y_m) < 2 \epsilon\]The above inequalities imply that: \(\vert d(x_m,y_m) - d(x_n,y_n) \vert < 2 \epsilon \\ d[d(x_m,y_m) - d(x_n,y_n)] < 2 \epsilon\)

This implies that \(a_n=d(x_n,y_n)\) is Cauchy, and thus converges.

\[\blacksquare\]**Lemma 1.4-2(b)** is: Let \(X=(X,d)\) be a metric space. Then, if \(x_n \rightarrow x\) and \(y_n \rightarrow y\), then \(d(x_n,y_n) \rightarrow d(x,y)\).

**Proof:**

We have \(x_n \rightarrow x\) and \(y_n \rightarrow y\). Then \((x_n)\) and \((y_n)\) are Cauchy. Thus the following two statements hold true:

- \(\forall \epsilon/2>0, \exists M\) such that \(d(x_m,x_n)<\epsilon/2\) for \(m,n>M\)
- \(\forall \epsilon/2>0, \exists N\) such that \(d(y_m,y_n)<\epsilon/2\) for \(m,n>N\)

Taking \(N_0=\max(M,N)\), the above statements become:

\(\forall \epsilon/2>0, \exists N_0\) such that \(d(x_m,x_n)<\epsilon/2\) and \(d(y_m,y_n)<\epsilon/2\) for \(m,n>N_0\), i.e., \(d(x_m,x_n)+d(y_m,y_n)<\epsilon/2+\epsilon/2=\epsilon\)

We will prove the result using proof by contradiction.

Suppose \((a_n)=(d(x_n,y_n))\) Suppose the claim is not true. Then, \(\require{cancel} a_n \cancel\rightarrow a\), thus \((a_n)\) is not Cauchy. This implies that: \(\exists \epsilon\) such that \(\forall N\), we have \(d(a_m, a_n)>\epsilon\) for all \(m,n>N\).

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

This is then true for arbitrary \(\epsilon\). But, this implies that for all \(N\), we cannot make \(d(x_m,x_n)+d(y_m,y_n)<\epsilon\). This is a contradiction, since by assumption, we have

\[d(x_m,x_n)+d(y_m,y_n)<\epsilon\]Thus \((a_n)\) is Cauchy, and is thus a convergent sequence.

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

We wish to show that if \(L_1\) is the limit of a Cauchy sequence in \((X,d_1)\) (call it \(x_n(d_1)\))and \(L_2\) is the limit of a Cauchy sequence in \((X,d_2)\) (call it \(x_n(d_2)\)), then \(L_1=L_2\).

We have \(\forall x, y \in X\), \(a.d_1(x,y) \leq d_2(x,y) \leq b.d_1(x,y)\).

Then, the **Triangle Inequality** gives us:

Applying the given metric constraints:

\[d_1(L_1,L_2) \leq d_1(L_1,x) + \frac{1}{a} d_2(x,L_2)\]We know that \(x_n(d_1) \rightarrow L_1\) and \(x_n(d_2) \rightarrow L_2\), therefore. If we have the following:

- \(\forall \epsilon>0, \exists M\) such that \(d_1(x_m(d_1),L_1)<\epsilon\) for \(m>M\)
- \(\forall \epsilon>0, \exists N\) such that \(d_1(x_m(d_2),L_2)<\epsilon\) for \(m>N\)

Pick \(N_0=\max(M,N)\), so that the above holds true for \(N_0\).

Then \(d_1(L_1,x_{N_0})<\epsilon\) and \(d_2(L_2,x_{N_0})<\epsilon\), so that we get:

\[d_1(L_1,L_2) < \epsilon \left(1+\frac{1}{a} \right)\]Since the above is true for all \(\epsilon>0\), we can conclude that \(d_1(L_1,L_2)=0\). Hence \(L_1=L_2\).

The same procedure can also be showing using \(d_2\).

\[\blacksquare\]The three distance metrics mentioned are:

- \[d_1(x,y)=d(x_1,x_2)+d(y_1,y_2)\]
- \[d_2(x,y)=\sqrt{ {d(x_1,x_2)}^2+{d(y_1,y_2)}^2}\]
- \[d_{max}(x,y)=\max [d(x_1,x_2),d(y_1,y_2)]\]

**Proof:**

For \(d_1\) and \(d_2\), let’s determine the conditions.

\[x+y \leq \sqrt{x^2+y^2} \\ x^2+y^2+2xy \leq x^2+y^2\]This gives us \(2xy \leq 0\), so that’s invalid; if we however we introduce a \(\sqrt{2}\) on the right hand side, we get:

\[x+y \leq \sqrt{2(x^2+y^2)} \\ x^2+y^2+2xy \leq 2x^2+2y^2 \\ x^2+y^2-2xy \geq 0\]which works. For the reverse inequality \(x+y \geq \sqrt{x^2+y^2}\), note that we immediately get \(2xy>0\), which works, so we can write the combined inequalities as:

\[\sqrt{x^2+y^2} \leq x+y \leq \sqrt{2} \sqrt{x^2+y^2} \\ \Rightarrow d_2(x,y) \leq d_1(x,y) \leq \sqrt{2} d_2(x,y)\]For \(d_1\) and \(d_max\), note that \(x+y> \geq \max(x,y)\) and \(2 \max(x,y) \geq x+y\)

Then, we get:

\[\max(x,y) \leq x+y \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_1(x,y) \leq 2 d_{ max}(x,y)\]For \(d_2\) and \(d_max\), note that \(x^2+y^2> \geq {\max(x,y)}^2\) and \(2 {\text{ max }(x,y)}^2 \geq x^2+y^2\)

Then, we get:

\[\max(x,y) \leq \sqrt{x^2+y^2} \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_2(x,y) \leq 2 d_{max}(x,y)\] \[\blacksquare\]**Proof:**

Assume \(\mathbb{R}\) is complete.

Assume two Cauchy Sequences in \(\mathbb{R}\):

- \((x_n)=x_1,x_2,\cdots\) converges to \(x\).
- \((y_n)=y_1,y_2,\cdots\) converges to \(y\).

Construct a sequence in \(\mathbb{C}\), like so:

\[(z_n)=x_1+iy_1,x_2+iy_2,\cdots\]Assume the distance metric for \(\mathbb{Z}\) is \(d(z_1,z_2)=\sqrt{ {(x_1-x_2)}^2 + {(y_1-y_2)}^2}\).

- \(\forall \epsilon>0, \exists M\) such that \(x_m-x<\frac{\epsilon}{\sqrt{2}}\) for \(m>M\)
- \(\forall \epsilon>0, \exists N\) such that \(x_m-x<\frac{\epsilon}{\sqrt{2}}\) for \(m>N\)

Pick \(N_0=\max(M,N)\), so that the above holds true for \(N_0\).

Pick \(z_i\) so that \(i>N_0\). Assume \(z=x+iy\). Then, we have:

\[d(z_i,z)=\sqrt{ {(x_i-x)}^2 + {(y_i-y)}^2}<\epsilon\]Then, for an arbitrary \(\epsilon\), there exists \(N_0\), such that \(d(z_i,z)<\epsilon\). Furthermore \(z \in \mathbb{C}\). Thus, \(\mathbb{C}\) contains the limits of all its Cauchy sequences. Thus, it is a closed set; hence it is a complete metric space.

\[\blacksquare\]tags: