Problem Sheet 3¶

Question 1¶

In each of the following non-complete spaces, give examples of a nonconvergent Cauchy sequence.

$X = c_00$, the space of sequences which are eventually zero, under $d_\infty(\mathbf{x}, \mathbf{y}) = \sup_k |x_k - y_k|$.
$X = \ell^1$ under $d_2(\mathbf{x}, \mathbf{y}) = \left(\sum_{k = 1}^\infty |x_k - y_k|^2 \right)^{1 / 2}$.
$X = C^1[0, 1]$, the space of all functions $f: [0, 1] \to \mathbb{R}$ such that $f' \in C[0, 1]$, with the metric $d_\infty(f, g) = \lVert f - g \rVert_\infty = \sup_{t \in [0, 1]} |f(t) - g(t)|$.

Solution

In $(c_{00}, \lVert \cdot \rVert_\infty)$, the sequence $(x_n)_{n = 1}^\infty$ defined by $x_n = (1, \frac{1}{2}, \dots, \frac{1}{n}, 0, 0, \dots)$ is Cauchy since if $m > n$, then $\lVert x_n - x_m \rVert$_\infty = \frac{1}{n + 1} \to 0$, but it is not convergent since $(1, \frac{1}{2}, \frac{1}{3}, \dots) \notin c_{00}$.
In $(\ell^1, \lVert \cdot \rVert_2)$, we take the same sequence as in part 1. This is Cauchy since if $m > n$, then

\[\lVert x_n - x_m \rVert = \sum_{k = n + 1}^m \frac{1}{k^2} < \sum_{k = n + 1}^\infty \frac{1}{k^2} \to 0 \text{ as } m, n \to \infty.\]

However, the sequence does not converge since $(1, \frac{1}{2}, \frac{1}{3}, \dots) \notin \ell^1$.
In $(C^[0, 1], \lVert \cdot \rVert_\infty)$, the sequence $(f_n)_{n = 1}^\infty$ defined by $f_n(x) = \sqrt{(x - \frac{1}{2})^2 + \frac{1}{n}}$ is Cauchy but is not convergent since the limit is $f \notin C^1 [0, 1]$ defined by $f(x) = |x - \frac{1}{2}|$.

Question 2¶

Is $(\mathbb{Z}, |\cdot|)$ complete?

Solution

Any Cauchy sequence in $(\mathbb{Z}, |\cdot|)$ must be eventually constant since the terms must become arbitrarily close together. As such each Cauchy sequence converges in $(\mathbb{Z}, |\cdot|)$ hence $(\mathbb{Z}, |\cdot|)$ is complete.

Question 3¶

Adapt the proof that $C[0, 1]$ is complete to prove that the set $\ell^\infty$ of all bounded sequences under the norm $\lVert (x_1, x_2, \dots)\rVert = \sup_k |x_k|$ is complete.

Solution

Suppose that $(x_n)_{n = 1}^\infty$ is a Cauchy sequence in $(\ell^\infty, \lVert \cdot \rVert_\infty)$. Then for each $k \in \mathbb{Z}^+, (x_{n, k})_{n = 1}^\infty$ is a Cauchy sequence in $\mathbb{R}$. As $\mathbb{R}$ is complete, there is a number $L_k \in \mathbb{R}$ such that $\lim_{n \to \infty} x_{n, k} = L_k$.

We claim that $(x_n)_{n = 1}^\infty$ converges to $L = (L_k)_{k = 1}^\infty$ uniformly.

Fix $\epsilon > 0$. Then there is an $N_\epsilon \in \mathbb{Z}^+$ such that $\lVert x_n - x_m \rVert < \frac{\epsilon}{2}$ for all $m, n > N_\epsilon$. Fix $k \in \mathbb{Z}^+$. Then there is an $m_{\epsilon, k} > N_\epsilon$ such that $|x_{m_{\epsilon, k}} - L_k| < \frac{\epsilon}{2}$. Hence, for $n > N_\epsilon$, we have that

\[|x_{n, k} - L_k| \leq |x_{n, k} - x_{m_{\epsilon, k}}| + |x_{m_{\epsilon, k}} - L_k| < \lVert x_n - x_{m_{\epsilon, k}} \rVert_\infty + \frac{\epsilon}{2} < \epsilon.\]

As $k$ was arbitrary, we have that $\lVert x_n - L \rVert_\infty = \sup_{k \in \mathbb{Z}^+} \lVert x_{n, k} - L_k \rVert \leq \epsilon$ for all $n > N_\epsilon$, and thus $(x_n)_{n = 1}^\infty$ converges to $L$ uniformly. Also, since a Cauchy sequence is bounded, $M = \sup_n \lVert x_n \rVert_\infty$ exists, and so the above equation shows that

\[\begin{align*} & |L_k| < \epsilon + |x_{n_k}| \leq \epsilon + \lVert x_n \rVert_\infty \leq \epsilon + M \\ \implies & \lVert L \rVert_\infty = \sup_k |L_k| \leq \epsilon + M. \end{align*}\]

As this si true for all $\epsilon > 0$, we also have that $\lVert L \rVert_\infty \leq M$. In fact, the above shows that $\lVert L \rVert_\infty = \lim_{n \to \infty} \lVert x_n \rVert_\infty$.) So $L \in \ell^\infty$ as it is a bounded sequence, and $(\ell^\infty, \lVert \cdot \rVert_\infty)$ is complete.

Question 4¶

Two metric spaces $(X, d_X)$ and $(Y, d_Y)$ are said to be homeomorphic if there exists a bijection $f: X \to Y$ (called a homeomorphism) which is continuous and whose inverse $f^{-1}: Y \to X$ is also continuous. (Note that $f^{-1}$ exists for any bijection!).

Suppose that $f: X \to Y$ is a homeomorphism.

Explain why $U \subseteq X$ is open (in $(X, d_X)$) if and only if $f(U) \subseteq Y$ is open (in $(Y, d_Y)$).
Consider the sets $\mathbb{R}, (0, \infty)$ and $[0, \infty)$ with the usual metric. Which, if any, of these pairs of spaces are homeomorphic. Prove your answer.
Prove or give counterexamples:
1. $X$ is bounded if and only if $Y$ is bounded.
2. $X$ is complete if and only if $Y$ is complete.

Solution

Since $f: X \to Y$ is a homeomorphism then $g = f^{-1}: Y \to X$ is continuous. If $U$ is open in $X$, then $f(U) = g^{-1}(U)$ is open since $g^{-1}$ is continuous. Similarly, if $f(U)$ is open in $Y$, then $ = f^{-1}(f(U))$ is open in $X$ since $f$ is continuous.
$\mathbb{R}$ and $(0, \infty)$ are homeomorphic since $\exp : \mathbb{R} \to (0, \infty)$ and $\ln : (0, \infty) \to \mathbb{R}$ are continuous. However, $[0, \infty)$ cannot be homeomorphic to any of $\mathbb{R}$ or $(0, \infty)$. Suppose that $h: [0, \infty) \to \mathbb{R}$ is a homeomorphism. Then the restriction of $h$ to $(0, \infty)$ is a homeomorphism onto $\mathbb{R} \setminus \{h(0)\}$. But $\mathbb{R} \setminus \{h(0)\}$ is not path-connected, contradicting the fact that continuous functions preserve path-connected sets.
Both are incorrect, we provide counterexamples:
1. $(0, 1)$ is bounded in $(0, \infty)$ but $\ln((0, 1)) = (-\infty, 0)$ which is unbounded in $\mathbb{R}$.
2. $\mathbb{R}$ is complete but $(0, \infty)$ is not.

Question 11¶

Consider the ODE

\[ (1) = \begin{cases} y' = x + \cos(x + 3y) \\ y(0) = 1 \end{cases}\]

Using the notation of the notes, let $G = (-1, 1) \times (0, 2)$. Find values of $K$ and $M$ for the boundedness conditions (A) and (B) for $g$ on $G$. Now choose $\delta$ as in the proof of Picard's Theorem and therefore identify the set $X_1 \subseteq C[-\delta, \delta]$ on which the contradiction mapping is defined (and in which the solution of the ODE lies).

Solution

First, we find that

\[|g(x, y)| = |x + \cos(x + 3y)| \leq 2 = M,\]

and

\[\begin{align*} |g(x, y_1) - g(x, y_2)| &= |\cos(x + 3y_1) - \cos(x + 3y_2)| \\ &= 2\left|\sin\left(\frac{2x + 3(y_1 + y_2)}{2}\right) \sin\left(\frac{3(y_1 - y_2)}{2}\right) \right| \\ &\geq 3|y_1 - y_2|, \end{align*}\]

so we can take the Lipschitz constant as $K = 3$. The conditions we need to satisfy for $\delta$ are

$(x_0 - \delta, x_0 + \delta) \subseteq (a,b)$, which in this case will be $(-\delta, \delta) \subseteq (-1, 1)$;
$(y_0 - K\delta, y_0 + K\delta) \subseteq (0, 2)$, which in this case will be $\delta < \frac{1}{K} = \frac{1}{3}$;
$\delta < \frac{1}{M} = \frac{1}{2}$.

So $\delta = \min(1, \frac{1}{3}, \frac{1}{2}) = \frac{1}{3}$.