Problem Sheet 6¶

Question 1¶

Consider the following nets. Do they converge?

$A = \mathbb{R}$ with the usual (partial) order $\leq$. The net is ${ \frac{1}{1 + x^2} }_{x \in A}.
$A = \mathbb{Z}^+ \times \mathbb{Z}^+$ with partial order

\[(n_1, m_1) \preceq (n_2, m_2) \iff n_1 \leq n_2 \text{ and } m_1 \leq m_2. \]

The net is $x_{n, m} = \frac{nm}{n^2 + m^2}, (n, m) \in A$.
The same net but with order

\[(n_1, m_1) \trianglelefteq (n_2, m_2) \iff n_1 < n_2 \text{ or } (n_1 = n_2 \text{ and } m_1 \leq m_2). \]

Solution

Yes converges to 0.
Given any $(n_0, m_0) \in A$, we have that $(n_0, m_0) \preceq (N, N)$ and $x_{N, N} = \frac{1}{2}$ for all $N \geq \max(n_0, m_0)$. But for any $n > 2m_0$, we have that $(n_0, m_0) \preceq (n, m_0)$ and

\[x_{n, m_0} = \frac{m_0}{n + m_0^2 / n} < \frac{m_0}{n} < \frac{1}{2}.\]
Does not converge with similar reasons to part 2.

Question 3¶

True of false? Suppose that $\tau_1$ and $\tau_2$ are two topologies on $X$ with the property that whenever $x_\alpha \to x$ in the $\tau_1$ topology then $x_\alpha \to x$ in the $\tau_2$ topology. Then $\tau_2 \subseteq \tau_1$.

Solution

True. Suppose that $\tau_2 \nsubseteq \tau_1$. Then there is some $U \in \tau_2 \setminus \tau_1$. Now,

\[U \supset S := \bigcup_{A \in \tau_1, A \subset U} A \in \tau_1, \]

otherwise $U$ would be open in $\tau_1$. Let $x \in U \setminus S$. Now, let $V$ be an open neighbourhood of $x$ with respect to $\tau_1$. Then $V \setminus U \neq \emptyset$ by the choice of $x$ (otherwise we would have $V \subset U$ and $x \in S$), and we let $x_V \in V \setminus U$. Then, considering the collection of open neighbourhoods of $x$ in $\tau_1$, denoted by $\mathcal{O}_1(x)$, ordered by reverse direction, we have that $\{x_V\}_{V \in \mathcal{O}_1(x)}$ is a net the converges to $x$ in $\tau_1$ but does not converge to $x$ in $\tau_2$.

Question 4¶

Consider $\mathbb{R}$ with the cofinite topology.

Describe what it means to be a neighbourhood of the number 2.
Let $\Lambda = \mathrm{Nbhd}(2)$ be the directed set of neighbourhoods of 2 with direction given by $V \leq W$ iff $W \subseteq V$. Given an example of a net indexed by $\Lambda$ with values in $[0, 1]$ which converges to 2.

Solution

The set $V$ is a neighbourhood of 2 if $2 \in V$ and $\mathbb{R} \setminus V$ is finite.
Let $V$ be a neighbourhood of 2 and take $x_V \in V \cap [0, 1]$. Then the net $\{x_V\}_{V \in \mathrm{Nbhd}(2)}$ takes its values in $[0, 1]$ and converges to 2. Indeed, let $V_0$ be a neighbourhood of 2. Then for any $V \geq V_0$ (so that $V \subseteq V_0$), we have that $x_V \in V \cap [0, 1] \subseteq V_0$. That is, the net is always eventually contained in any neighbourhood of 2.

Question 5¶

Give an example of a continuous bijection between two topological spaces which is not a homeomorphism.
Show that any continuous bijection between $\mathbb{R}$ and itself (with the usual topology in both domain and codomain) is a homeomorphism.

Solution

$f: [0, 2\pi) \to \mathbb{T} = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\}$ defined by $f(\theta) = \exp(i\theta)$ is a continuous bijection but which is not a homomorphism since $f^{-1}$ is not continuous, as $f([0, \pi)) is not open in $\mathbb{T}$.
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous bijection with respect to the standard topologies. Then $f$ must be strictly increasing or decreasing and without loss of generality, we assume that it is increasing. Then $g = f^{-1}$ is also a strictly increasing function. Let $a \in \mathbb{R}$ and $\epsilon > 0$. Take $b$ and $c$ such that $g(b) = g(a) - \epsilon$ and $g(c) = g(a) + \epsilon$. Then $b < a < c$ and let $\delta = \min\{a - b, c - a\}$ so that if $|x - a| < \delta$, we have that $|g(x) - g(a)| < \epsilon$.

Question 6¶

Let $Y$ be the set of all $2 \times 2$ matrices with determinant 1. This is a subset of the set of all $2 \times 2$ matrices which is a normed space under several equivalent norms. If you wish, you can take

\[\lVert \begin{pmatrix} a & b \\ c & d \end{pmatrix}\rVert = |a| + |b| + |c| + |d|. \]

Find a continuous path in $X$ joining

\[\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \text{ to } \quad \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix}. \]

Solution

$F: [0, 1] \to \mathrm{SL}_2(\mathbb{R})$ defined by

\[ F(t) = \begin{pmatrix} \cos 2\pi t & -\sin 2\pi t \\ \sin 2\pi t & \cos 2\pi t \end{pmatrix}\]

Question 7¶

Let $Y$ be the set of all invertible functions in $C[0, 1]$. (By that we mean that $1/f$ needs to be in $C[0, 1]$.) Show that $Y$ is connected if we are using complex scalars, but not if we are using real scalars.

Solution

With real scalars, there is no path from 1 to -1. Suppose that there is such a path $F: [0, 1] \to Y$. Then $t \mapsto F(t)(1)$ defines a continuous function on $[0, 1]$ with $F(0)(1) = 1$ and $F(1)(1) = -1$. By the Intermediate Value Theorem, there is some $c$ such that $F(c)(1) = 0$, but this means that $F(c)$ is not an invertible function.

Showing that $Y$ is path connected under complex scalars is tricky. One could try to find a path from any invertible function $f$ to 1, but this requires a fact which is that the argument function for $f$ is continuous...

Question 8¶

Show that if $(X, \tau)$ is a topological space and $\{V_i\}_{i \in I}$ is a collection of connected subsets of $X$ such that $\bigcap_{i \in I} V_i \neq \emptyset$, then $\bigcup_{i \in I} V_i$ is connected.

Solution

Suppose that $U_1, U_2$ are disjoint open subsets of $X$ such that $\bigcup_{i \in I} V_i \cap (U_1 \cup U_2) = \bigcup_{i \in I} V_i$, and let $x \in \bigcap_{i \in I} V_i$. Without loss, assume that $x \in U_1$. We aim to show that $\bigcup_{i \in I} V_i \cap U_2 = \emptyset$. For each $i \in I$, we have that $V_i \cap (U_1 \cup U_2) = V_i$. As $x \in V_i \cap U_1$ and $V_i$ is connected, we have that $V_i \cup U_2 = \emptyset$ for every $i \in I$. Thus $(\bigcup_{i \in I} V_i) \cap U_2 = \bigcup_{i \in I} V_i \cap U_2 = \bigcup_{i \in I} \emptyset = \emptyset$. So $\bigcup_{i \in I} V_i$ is connected.

Question 9¶

Give an example of a topological space which has infinitely many connected components, none of which are open sets.

Solution

$\mathbb{Q}$ with the standard topology.

Question 10¶

Show that if $(X, \tau)$ is a topological space, and $Y \subseteq X$ is connected, then $\mathrm{cl}(Y)$ is connected as well.

Solution

Suppose that $U_1$ and $V_1$ are disjoint open subsets of $X$ such that $\mathrm{cl}(Y) = \mathrm{cl}(Y) \cap (U_1 \cup U_2)$. Then $Y = Y \cap (U_1 \cup U_2)$ and, as $Y$ is connected, we can assume that $Y \cap U_2 = \emptyset$ and $Y \cap U_1 = Y$. Suppose that $x \in \mathrm{Bd}(Y) \cap U_2$. As $U_2$ is an open neighbourhood of $x \in \mathrm{Bd}(Y)$, then it should be that $Y \cap U_2 \neq \emptyset$, but this is a contradiction. So $\mathrm{Bd}(Y) \cap U_2 = \emptyset$, and thus $\mathrm{cl}(Y) \cap U_2 = \emptyset$.

Question 11¶

Let $X = \{(x, \sin \frac{1}{x})\}_{x \in (0, 1)} \cap \{(0, 0)\}$, with the topology inherited from $\mathbb{R}^2$. Show that $X$ is connected but not path-connected.

Solution

It is easy to see that it is not path connected, since any continuous function from $(1, \sin(1))$ to $(-1, \sin 1)$ would have to satisfy $F(t_n) = (\frac{2}{(4n + 1)\pi}, 1)$ but with $\lim_{n \to \infty} t_n = 0$ but $F(t_n) \to (0, 1)$.

Suppose that $U_1 \cup U_2$ are disjoint open sets in $\mathbb{R}^2$ such that $X \cap (U_1 \cup U_2) = X$, and without loss we assume that $(0, 0) \in U_1$. Then $U_1$ contains an open ball centred at $(0, 0)$, say $B$. Then $X \cap B$ can be joined with two other connected sets to make $X$, and since the union of two connected sets with non-empty intersection is connected, we have that $U_2 \cap X = \emptyset$.