Problem Sheet 5¶

Question 1¶

Let $X = \mathbb{R}$. Define

\[ \tau = \{\emptyset, \mathbb{R}\} \cup \{(-t, t): t > 0\}. \]

Prove that $(X, \tau)$ is a topological space.
For $k = 1, 2, \dots$ let $x_k = \sin(k!)$. Prove that the sequence $\{x_k\}$ converges to $\frac{\pi}{e}$ in this topology.
Characterise the sequences which converge to 0, and those which converge to 1.

Solution

Suppose that $\{U_i\}_{i \in I} \subseteq \tau$. If $U_i = \emptyset$ for all $i$, then $\bigcup_{i \in I} U_i = \emptyset \in \tau$. If $\mathcal{U}_i = \mathbb{R}$ for some $i \in I$, then $\bigcup_{i \in I} U_i = \mathbb{R} \in \tau$. Otherwise, without loss of generality (by omitting any of the $U_i$ that are empty), suppose that for each $i \in I$ that $U_i = (-t, t_i)$ for some $t_i \in \mathbb{R}^+$. If $\{t_i\}_{i \in I}$ is unbounded, then $\bigcup_{i \in I} U_i = \mathbb{R} \in tau$. Otherwise, $s = \sup_{i \in I} t_i$ exists, and $\bigcup_{i \in I} U_i = (-s, s) \in \tau$. So $\tau$ is closed under arbitrary unions.

Now suppose that $\{U_i\}_{i = 1}^n \subseteq \tau$. If $U_i = \emptyset$ for any $i$, then $\bigcap_{i = 1}^n U_i = \emptyset \in \tau$. If $U_i = \mathbb{R}$ for all 4i$, then $\cap_{i = 1}^n U_i = \mathbb{R} \in \tau$. So suppose without loss of generality that $U_i = (-t_i, t_i)$ for all $i$. Then $\cap_{i = 1}^n U_i = (-m, m)$ where $m = \min_{1 \leq i \leq n} t_i$. So $\tau$ is closed under finite intersections, and is also a topology on $\mathbb{R}$. (Note that is suffices to consider just two sets since induction will extend the result to finitely many sets.)
For each $k \in \mathbb{Z}^+, \sin(k!) \in [-1, 1] \subseteq (-t, t)$ for any $t > \pi / e > 1$. Now, all the open neighbourhoods of $\pi / e$ are of the form $(-t, t)$ for some $t > \pi / e$. So $\{\sin(k!)\}_{k = 1}^\infty$ is contained in every neighbourhood of $\pi / e$, and thus converges to $\pi / e$ with respect to $\tau$.
A sequence converges to 0 with respect to $\tau$ if and only if it converges to 0 in the standard topology:

\[ x_k \xrightarrow{\tau} 0 \iff \forall t > 0 \exists K \in \mathbb{Z}^+ \forall k > K x_k \in (-t, t) \iff x_k \xrightarrow{\text{std}} 0. \]

Claim. A sequence $\{x_k\}_{k = 1}^\infty$ converges to 1 with respect to $\tau$ if and only if $\limsup_{k \to \infty} |x_k| \leq 1$.

Suppose that $x_k \xrightarrow{\tau} 1$. For any $\epsilon > 0$, there is a $K_\epsilon \in \mathbb{Z}^+$ such that $x_k \in (-1 - \epsilon, 1 + \epsilon)$ for all $k > K_\epsilon$. That is, $|x_k| < 1 + \epsilon$ for all $k > K_\epsilon$. Hence $\limsup_{k \to \infty} |x_k| \leq 1$.

Now suppose that $\limsup_{k \to \infty} |x_k| \leq 1$. Then for any $t > 1$, there is a $K_t \in \mathbb{Z}^+$ such that $|x_k| < t$ for all $k > K_t$. That is, $x_k \in (-t, t)$ for all $k > K_t$, and so $x_k \xrightarrow{\tau} 1$.

Question 2¶

Which of the following are topologies? What topologies are Hausdorff?

$X = \mathbb{R}. \tau = \{ \emptyset, \mathbb{R}\} \cup \{(-\infty, a) : a \in \mathbb{R}\}.$
$X = \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R}$ is a fixed function. $\tau = \{f^{-1}(U): U \text{ is an open set in } \mathbb{R} \text{ in the usual topology}\}.$
$X = D = \{z \in \mathbb{C}: |z| \leq 1\}$. $\tau = \{ U \cap D : U \text{ is an open set in } \mathbb{C} \text{ in the usual topology}\}$.
$X = C[0, 1].$ For $r > 0$ let $B_r = \{f \in C[0, 1] : \lVert f \rVert_\infty < r\}$. $\tau = \{\emptyset, X\} \cup \{B_r : r > 0\}$.
$X$ be any set. $\tau = \{\emptyset\} \cup \{ U \subset X : X \setminus U \text{ is finite}\}$.

Solution

This is a topology however is not Hausdorff. If $x, y \in \mathbb{R}$ with $x < y$ then any open neighbourhood containing $y$, will also be an open neighbourhood of $x$.
Firstly $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(\mathbb{R}) = \mathbb{R}$. If $\{f^{-1}(U_i)\}_{i \in I} \subseteq \tau_f$, then since $\bigcup_{i \in I} U_i$ is open and $f^{-1}(\bigcup_{i \in I} U_i) = \bigcup_{i \in I} f^{-1}(U_i)$, we have that $\tau_f$ is closed under arbitrary unions. A similar argument will show that $\tau_f$ is closed under finite intersections, and so $\tau_f$ is a topology (that makes $f$ continuous!).

We claim that $\tau_f$ is Hausdorff if and only if $f$ is injective.

First suppose that $f$ is injective and let $x, y \in \mathbb{R}$ be distinct, so that $f(x) \neq f(y)$. As the standard topology is Hausdorff, we can find disjoint open sets $U_x \ni f(x)$ and $U_y \ni f(y)$. Then $f^{-1}(U_x) \ni x$ and $f^{-1}(U_y) \ni y$ are disjoint and both in $\tau_f$, and so $\tau_f$ is Hausdorff.

Conversely, suppose that $f$ is not injective. Then there are distinct $x, y \in \mathbb{R}$ such that $f(x) = f(y)$. If $x \in f^{-1}(U)$ for some open $U$, then $f(X) = f(y) \in U$, and so $y \in f^{-1}(U)$. So any open neighbourhood of $x$ also contains $y$, and thus $\tau_f$ is not Hausdorff.
This is just the subspace/relative topology.
This defines a non-Hausdorff topology. Non-Hausdorff since every non-empty open set contains 0. You can show union and intersections by taking $\sup$ of the radii.
This is the cofinite topology, is not Hausdorff.

Question 3¶

Suppose that $(X, d)$ is a metric space and that $Y$ is a nonempty subset of $X$. As usual, the metric determines a topology $\tau$ on $X$, so we can talk about subspace/relative topology $\tau_Y$ induced by $\tau$ on $Y$. On the other hand, we can consider $(Y, d)$ as a metric space, and so $d$ also directly determines a topology $\tau_Y'$ on $Y$. Check that these two topologies are the same.

Solution

Suppose that $U$ is in the subspace topology on $Y$, so that $U = V \cap Y$ for some open $V$ in $X$. Let $y \in U$ so that there is some $\epsilon > 0$ with $B_X(y, \epsilon) \subset V$. Then $B_Y(y, \epsilon) = B_X(x, \epsilon) \cap Y \subseteq U$, and so every point of $Y$ is an interior point of $Y$ with respect to the induced metric. That is, $U$ is open in the induced metric topology.

Conversely, suppose that $U$ is open in the induced metric topology on $Y$. Then for each $y \in U$, we can find an $\epsilon_y > 0$ such that $B_Y(y, \epsilon_y) \subseteq U$, and so $U = \bigcup_{y \in U} B_Y(y, \epsilon_y) = \left(\bigcup_{y \in U} B_X(y, \epsilon_y)\right) \cap Y$, which is an element of the subspace topology on $Y$.

Question 4¶

If $Y \subseteq (X, \tau)$, prove that a sequence $\{y_n\}_{n = 1}^\infty \subseteq Y$ converges in the relative topology to an element $y \in Y$ if and only if it converges in $(X, \tau)$.

Solution

Suppose that $(y_n)_{n = 1}^\infty$ is a sequence in $Y$ that converges to $y \in Y$ with respect to the subspace topology. Now let $U \subseteq X$ be an open neighbourhood of $y$. Then there is an $N_u \in \mathbb{N}$ such that $y_n \in U \cap Y \subseteq U$ for all $n > N_U$. That is, $y_n \to y$ in $X$ with respect to $\tau$.

Conversely, suppose that $y_n \to y \in Y$ with respect to $\tau$. Let $U \cap Y$ be an open neighbourhood of $y$ in the relative topology. Then there is an $N_u \in \mathbb{N}$ such that $y_n \in U$ for all $n > N_U$. Since the sequence is in $Y$, we have that $y_n \in U \cap Y$ for all $n > N_U$. Thus $y_n \to y$ in the subspace topology on $Y$.

Question 5¶

Define a base for the ``line with two origins'' topology.

Solution

The line with two origins is $\mathbb{R} \setminus \{0\} \cup \{0_a, 0_b\}$. A base for its topology would be

\[ \{(a, b) : a < b < 0 \text{ or } 0 < a < b\} \cup \{(-\epsilon, 0) \cup \{0_a\} \cup (0, \epsilon), (-\epsilon, 0) \cup \{0_b\} \cup (0, \epsilon)\}_{\epsilon > 0} \]

Clearly this set covers the space, and if two of these sets have non-empty intersection, then the intersection contains an interval.

Question 6¶

Which of the following is a base for a topology on $X$?

$X = \{a, b, c\}, \mathcal{B} = \{\{a, b\}, \{a, b, c\}\}$
$X = \{a, b, c\}, \mathcal{B} = \{\{a, b\}, \{b, c\}\}$
$X = \mathbb{R}^2, \mathcal{B}$ is a set of closed disks of positive radius.
$X = C[0, 1], \mathcal{B} = \{S_K\}_{K \in \mathbb{R}\}$

\[ S_k = \{f: |f(0)| \geq K\}. \]

Solution

Yes, with $\tau = \{\emptyset, \{a, b\}, X\}$.
No, since $b \in \{a, b\} \cap \{b, c\}$ but $\{ b\} \notin \mathcal{B}$
No, the intersection of two tangential disks will be a point, which does not contain a closed disk.
Yes, since it covers $C[0, 1]$ and $S_{k_1} \cap S_{k_2} = S_{\max(k_1, k_2)}$.

Question 7¶

Determine whether each of the topologies on $\mathbb{R}$ is first and/or second countable.

The usual topology.
The discrete topology.
The cofinite topology.

Solution

Second countable.
First but not second countable. A local base for $x \in \mathbb{R}$ would be $\{\{x\}\}$, but we need $\{\{x\}\}_{x \in \mathbb{R}}$ to be included in any base of the discrete topology.
Neither. Suppose that 0 has a countable local base $\{U_i\}_{i = 1}^\infty$. Then each $U_iu$ is of the form $\mathbb{R} \setminus F_i$ for some finite set $F_i \subset \mathbb{R}$. Let $x \in \cap_{i = 1}^\infty U_i \setminus \{0\}$, which is non-empty since $\mathbb{R}$ is uncountable and a countable union of finite sets is countable. Then $\mathbb{R} \setminus \{x\}$ is a neighbourhood of 0 and is not contained in any $U_i$. Thus $\mathbb{R}$ with the cofinite topology is not first countable, and hence is not second countable. (This argument will also apply to the cocountable topology).

Question 8¶

Show that if $(X, d)$ is a metric space and $Y$ is a dense subset, then $\{B(y, \frac{1}{n})\}_{y \in Y, n \in \mathbb{Z}_+}$ is a base for the metric topology.

Solution

Suppose that $U$ is an open subset in $(X, d)$ and let $u \in U$. Then there is some $n_u \in \mathbb{Z}^+$ such that $B(u, \frac{1}{n_u}) \subseteq U$. Since $Y$ is dense in $X$, we can find a $y_U \in Y$ such that $d(u, y_u) < \frac{1}{3n_u}$. Then we have that the ball $B(y_u, \frac{1}{2n_u})$ contains $u$, and is contained in $B(u, \frac{1}{n_u})$ since if $x \in B(y_u, \frac{1}{2n_u})$,

\[d(x, u) \leq d(x, y_u) + d(y_u, u) < \frac{1}{2n_u} + \frac{1}{3n_u} = \frac{5}{6n_u} < \frac{1}{n_u}. \]

So we can write $U = \bigcup_{u \in U} B(y_u, \frac{1}{n_u})$, and thus $\{B(y, \frac{1}{n})\}_{y \in Y, n \in \mathbb{Z}^+}$ is a base for the metric topology.

Question 9¶

Let $c_{00}(\mathbb{Q})$ be the set of sequences with rational coordinates which are eventually $0$. Show that for each $1 \leq p < \infty, c_{00}(\mathbb{Q})$ is dense in $\ell^p$.

Solution

Let $x \in \ell^p$ and $\epsilon > 0$. Then there is an $N \in \mathbb{Z}^+$ such that $\left(\sum_{n = N + 1}^\infty |x_n|^p\right)^{1 / p} < \frac{\epsilon}{n}$. Now, let $r_n \in \mathbb{Q}$ be such that $|x_n - x_r| < \frac{\epsilon}{2N^{1/p}}$ and define $r \in c_{00}(\mathbb{Q})$ by $r = (r_1, r_2, \dots, r_N, 0, 0, \dots)$. Then

\[ d_p(x, r) = \left(\sum_{n = 1}^\infty |x_n - r_n|^p\right)^{1 /p} < (N \left(\frac{\epsilon}{2N^{1/p}}\right)^{1/p} + \left(\sum_{n = N + 1}^\infty |x_n|^p \right)^{1/p} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]

Question 10¶

Show that $\ell^\infty$ is not separable.

Solution

Suppose that $S$ i a dense subset of $\ell^\infty$. For any $A \subseteq \mathcal{P}(\mathbb{N})$, let $e_A$ be the sequence defined by $(e_A)_n = 1$ if $n \in A$ and 0 otherwise. Then $\lVert e_A - e_B \rVert_\infty = 1$ for all $A \neq B$. For each $A$, we can find $s_A \in S$ such that $\lVert e_A - s_A \rVert_\infty < \frac{1}{2}$. However, for $A \neq B$, we have that

\[\begin{align*} \lVert s_A - s_B \rVert_\infty &= \lVert s_A - e_A + e_A - e_B + e_B - s_B \rVert_\infty \\ &\geq \lVert e_A - e_B \rVert_\infty - \lVert e_A - s_A \rVert_\infty - \lVert e_B - s_B \rVert_\infty \\ &> 1 - \frac{1}{2} - \frac{1}{2} = 0. \end{align*}\]

Thus we have an injection from $\mathcal{P}(\mathbb{N})$ into $S$, and $S$ is uncountable.