Exercises for Section 1.5

1. Let $B = \bigcup_{i \in I} A_i$, where each $A_i \in \mathbb{R}^n$ is open. Let $\mathbf{x} \in B$. Then there exists some index $i \in I$ s.t. $\mathbf{x} \in A_i$. Since $A_i$ is open, then there is an $r > 0$ s.t. $B_r(\mathbf{x}) \subset A_i \subset B$.

Hence for each point in $B$ there’s an open ball around it contained in $B$. Therefore, $B$ is open.

2. Let $B = \bigcap_{i \in I} A_i$ where $\lvert I\rvert = n \in \mathbb{N}$. Let $\mathbf{x} \in B$. Then for all $i \in I$, $\mathbf{x} \in A_i$, and since each $A_i$ is open, for all $i$ there exists $r_i > 0$ such that $B_{r_i}(\mathbf{x}) \subset A_i$. Let $p = \min_{i \in I} r_i$. Since $I$ is finite and every $r_i > 0$, we have $p > 0$. Then for all $i \in I$, $B_p(\mathbf{x}) \subset A_i$, which implies $B_p(\mathbf{x}) \subset B$.

Hence for each point in $B$ there’s an open ball around it contained in $B$. Therefore, $B$ is open.

3. Let $A_i=\left(-\dfrac{1}{2^i},\dfrac{1}{2^i}\right)$, which is open for each $i \in \mathbb{N}$. Define $B = \bigcap_{i=0}^{\infty} A_i$. Then $B = \\{0\\}$ which is not open.

1. If $X \subset A$ is an open set, then:

$$ \begin{align*} \mathbf{x} \in X &\implies \exists\,r > 0\quad\text{s.t.}\quad B_r(\mathbf{x}) \subset X \\ &\implies \exists\,r > 0\quad\text{s.t.}\quad B_r(\mathbf{x}) \subset A \\ &\implies \mathbf{x} \in \overset{\circ}{A} \end{align*} $$

Therefore, $X \subset \overset{\circ}{A}$. Also, $\overset{\circ}{A}$ itself is open and contained in $A$. So, $\overset{\circ}{A}$ is the largest open set contained in $A$.

2.

First we show that $\overline{A}$ is closed. To do this, we need to show that $\overline{A}^c$ is open. To do this, we need to show if $x \in \overline{A}^c$ then there is $r > 0$ s.t. $B_r(x) \subset \overline{A}^c$.

We’ll use the following lemma in our proof.

Lemma. If $B_r(\mathbf{x}) \cap A = \emptyset$, then $B_r(\mathbf{x}) \cap \overline{A} = \emptyset$

Using this lemma and the definition of closure, we have:

$$ \begin{align*} x \in \overline{A}^c &\implies x \notin \overline{A} \\ &\implies \neg (\forall r > 0 \quad \text{s.t.} \quad B_r(x) \cap A \ne \emptyset) \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(x) \cap A = \emptyset \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(x) \cap \overline{A} = \emptyset \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(x) \subset \overline{A}^c. \end{align*} $$

Therefore, $\overline{A}^c$ is open and $\overline{A}$ is closed.

Proof of Lemma. Assume $B_r(\mathbf{x}) \cap A = \emptyset$ but $B_r(\mathbf{x}) \cap \overline{A} \ne \emptyset$. Then there exists $\mathbf{y} \in B_r(\mathbf{x}) \cap \overline{A}$. Then by definition of open ball $r' = r - \lVert x - y\rVert > 0$, and by definition of closure $B_{r'}(\mathbf{y}) \cap A \ne \emptyset$. But $B_{r'}(\mathbf{y}) \subset B_{r}(\mathbf{x})$, which means $B_{r}(\mathbf{x}) \cap A \ne \emptyset$. A contradiction. So, such $\mathbf{y}$ doesn’t exist and $B_r(\mathbf{x}) \cap \overline{A} = \emptyset$.

Second, we show that $\overline{A}$ is the smallest closed set containing $A$.

Assume that there’s a smaller closed set $X$ such that $A \subset X$. Then there exists a $\mathbf{x} \in \overline{A}$ which is not in $X$. Since $X$ is closed, then $X^c$ is open and:

$$ \begin{align*} \mathbf{x} \in \overline{A} - X &\implies \mathbf{x} \in X^c \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(\mathbf{x}) \subset X^c \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(\mathbf{x}) \cap A \subset X^c \cap X \quad \text{(since } A \subset X \text{)} \\ &\implies \exists r > 0 \quad \text{s.t.} \quad B_r(\mathbf{x}) \cap A = \emptyset \\ &\implies \mathbf{x} \notin \overline{A} \end{align*} $$

Which is a contradiction. So, $\overline{A}$ is the smallest closed set containing $A$.

3. From problem 1.5.6 we have $\partial A = \overline{A} \cap \overline{A^c}$.

Then:

$$ \begin{align*} A \cup \partial A &= A \cup (\overline{A} \cap \overline{A^c}) \\ &= (A \cup \overline{A}) \cap (A \cup \overline{A^c}) \\ &= \overline{A} \cap (A \cup \overline{A^c}) \quad &\text{(since } A \subset \overline{A} \text{)} \\ &= \overline{A} \cap \mathbb{R}^n &\text{(since } A^c \subset \overline{A^c} \text{ and } A \cup A^c = \mathbb{R}^n \text{)} \\ &= \overline{A} \end{align*} $$

4.

By definition,

$$ \mathbf{x} \in \overset{\circ}{A} \Leftrightarrow \exists r > 0 \, B_r(\mathbf{x}) \subset A $$

Then:

$$ \mathbf{x} \in \left(\overset{\circ}{A}\right)^c \Leftrightarrow \forall r > 0 \, B_r(\mathbf{x}) \cap A^c \ne \emptyset $$

And since definition of closure is:

$$ \mathbf{x} \in \overline{A} \Leftrightarrow \forall r > 0 \, B_r(\mathbf{x}) \cap A \ne \emptyset $$

Then

$$ \mathbf{x} \in \overline{A} \cap \left(\overset{\circ}{A}\right)^c \Leftrightarrow \forall r > 0 \, B_r(\mathbf{x}) \cap A \ne \emptyset \text{ and } B_r(\mathbf{x}) \cap A^c \ne \emptyset $$

Which means

$$ \mathbf{x} \in \overline{A} - \overset{\circ}{A} \Leftrightarrow \mathbf{x} \in \partial{A}. $$

We need to show that $\mathbf{x} \in \partial A$ if and only if $\mathbf{x} \in \overline{A} \cap \overline{A^c}$

($\implies$) Assume $\mathbf{x} \in \partial A$. Then by definition of boundary we have:

1. $\forall r > 0$, $B_r({\mathbf{x}}) \cap A \ne \emptyset$, which implies $\mathbf{x} \in \overline{A}$.
2. $\forall r > 0$, $B_r({\mathbf{x}}) \cap A^c \ne \emptyset$, which implies $\mathbf{x} \in \overline{A^c}$.

Which means $\mathbf{x} \in \overline{A} \cap \overline{A^c}$.

($\impliedby$) Assume $\mathbf{x} \in \overline{A} \cap \overline{A^c}$. Then:

1. $\mathbf{x} \in \overline{A}$, which by definition means $\forall r > 0$, $B_r({\mathbf{x}}) \cap A \ne \emptyset$
2. $\mathbf{x} \in \overline{A^c}$, which by definition means $\forall r > 0$, $B_r({\mathbf{x}}) \cap A^c \ne \emptyset$

So, by definition of boundary, $x \in \partial A$.

1. Since $\lVert A^k \rVert \le \lVert A \rVert^k$, then $\sum_{k=0}^{\infty} \left\lVert\dfrac{A^k}{k!}\right\rVert \le \sum_{k=0}^{\infty} \dfrac{\lVert A\rVert^k}{k!}$, which by proposition 1.5.36 converges to $e^{\lVert A \rVert}$. Using proposition 1.5.35 (absolute convergence), we conclude $e^{A}$ itself converges.

To find a bound for $\lVert e^{A} \rVert$, we note that:

$$ \begin{align*} \lVert e^A \rVert &= \left\lVert\sum_{k=0}^{\infty} \frac{A^k}{k!}\right\rVert \\ &\le \sum_{k=0}^{\infty} \left\lVert\frac{A^k}{k!}\right\rVert \quad &\text{(exercise 1.5.9)} \\ &\le \sum_{k=0}^{\infty} \frac{\lVert A\rVert^k}{k!} \quad &(\lVert A^k \rVert \le \lVert A \rVert^k) \\ &= e^{\lVert A \rVert} \quad &\text{(proposition 1.5.36)} \end{align*} $$

Since norms are non-negative, then $0 \le \lVert e^A \rVert \le e^{\lVert A \rVert}$.

2. $e^{A+B} = e^{A}e^{B}$ is generally not true. For example, let $A = \begin{pmatrix} 0 & 1 \\\\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\\\ 1 & 0 \end{pmatrix}$. Then $A^2 = B^2 = \mathbf{0}$. So, $e^A = I + A = \begin{pmatrix} 1 & 1 \\\\ 0 & 1\end{pmatrix}$ and $e^B = I + B = \begin{pmatrix} 1 & 0 \\\\ 1 & 1 \end{pmatrix}$.

To calculate $e^{A+B}$, note that:

$$ (A + B)^k = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}^k = \begin{cases} \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} \quad \text{when k odd} \\ \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \quad \text{when k even} \end{cases} $$

So, $(A + B)^k = \begin{pmatrix} \dfrac{1 + (-1)^k}{2} & \dfrac{1 - (-1)^k}{2} \\\\ \dfrac{1 - (-1)^k}{2} & \dfrac{1 + (-1)^k}{2} \end{pmatrix}$, and

$$ \begin{align*} e^{A+B} &= \begin{pmatrix} \dfrac{1}{2}\sum_{k=0}^{\infty} \left(\dfrac{1}{k!} + \dfrac{(-1)^k}{k!}\right) & \dfrac{1}{2} \sum_{k=0}^{\infty} \left(\dfrac{1}{k!} - \dfrac{(-1)^k}{k!}\right) \\ \dfrac{1}{2} \sum_{k=0}^{\infty} \left(\dfrac{1}{k!} - \dfrac{(-1)^k}{k!}\right) & \dfrac{1}{2}\sum_{k=0}^{\infty} \left(\dfrac{1}{k!} + \dfrac{(-1)^k}{k!}\right) \end{pmatrix} \\ &= \begin{pmatrix} \dfrac{e+e^{-1}}{2} & \dfrac{e-e^{-1}}{2} \\ \dfrac{e-e^{-1}}{2} & \dfrac{e+e^{-1}}{2} \end{pmatrix} \end{align*} $$

So, $e^{A+B} = \begin{pmatrix} \cosh{1} & \sinh{1} \\\\ \sinh{1} & \cosh{1}\end{pmatrix} \ne \begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix} = e^A e^B$.

---

$e^{A+B} = e^{A}e^{B}$ holds when $AB = BA$.

We have:

$$ \begin{align*} e^{A+B} &= \sum_{k=0}^{\infty} \frac{(A+B)^k}{k!} \\ &= \sum_{k=0}^{\infty} \sum_{i=0}^{k} \frac{{k \choose i} A^iB^{k-i}}{k!} \quad \text{(\*)} \\ &= \sum_{k=0}^{\infty} \sum_{i=0}^{k} \frac{A^i B^{k-i}}{i! (k - i)!} \\ &= \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \frac{A^i B^j}{i! j!} \\ &= \left(\sum_{i=0}^{\infty} \frac{A^i}{i!}\right) \left(\sum_{j=0}^{\infty} \frac{B^j}{j!}\right) \\ &= e^Ae^B \end{align*} $$

Where we used $AB=BA$ in step (*).

---

$e^{2A} = \left(e^A\right)^2$ holds for all $A$ as a corollary of the previous result.

By definition it means:

$$ \forall \epsilon > 0 \exists \delta > 0\quad \text{s.t.} \quad \sqrt{x^2+y^2} < \delta \implies | f\begin{pmatrix} x \\ y \end{pmatrix} - a | < \epsilon $$

More intuitively it means: for any given $\epsilon$, there’s an open circle centered at the origin with radius $0 < \delta < 1$ such that value of $f$ for all points in that region is in the $(a - \epsilon, a + \epsilon)$ interval.

We have $\lim_{x\to0^-} f \begin{pmatrix} x \\\\ 0 \end{pmatrix} = \lim_{x\to 0^-}\dfrac{sin(x)}{-x} = -1$ and $\lim_{x\to0^+} f \begin{pmatrix} x \\\\ 0 \end{pmatrix} = \lim_{x\to 0^+}\dfrac{sin(x)}{x} = +1$.

So, the limit of this function at $\mathbf{0}$ doesn’t exist.

Let $r = \lVert (x, y) \rVert = \sqrt{x^2 + y^2}$ and $s = \lvert x \rvert + \lvert y \rvert$. Then $g = 2s \ln r$.

Using triangle inequality, we have $r \le s$, and using cauchy-schwarz inequality (with $\mathbf{u}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$, $\mathbf{v}=(|x|,|y|)$), we have $\dfrac{s}{\sqrt{2}} \le r$.

When $r < 1$, then $\ln r < 0$. So, near zero:

$$ \begin{align*} \dfrac{s}{\sqrt{2}} \le r \le s &\implies 2 \sqrt{2} \, r \ln r \le 2s \ln r \le 2r \ln r \\ &\implies 2 \sqrt{2} \, r \ln r \le g \begin{pmatrix} x \\ y \end{pmatrix} \le 2r \ln r \end{align*} $$

From one dimensional calculus we know $\lim_{r\to 0} r \ln r = 0$, which means:

$$ \lim_{\begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 0^+ \\ 0^+ \end{pmatrix}} g \begin{pmatrix} x \\ y \end{pmatrix} = 0 $$

We have

$$ A^k = \begin{pmatrix} \dfrac{(2a)^k}{2} & \dfrac{(2a)^k}{2} \\ \dfrac{(2a)^k}{2} & \dfrac{(2a)^k}{2} \end{pmatrix} $$

For $k \mapsto A^k$ to converge, $k \mapsto \dfrac{(2a)^k}{2}$ must converge. Then:

• If $|a| > \dfrac{1}{2}$, $A^k$ doesn’t converge.
• If $|a| < \dfrac{1}{2}$, $A^k$ converges to $\mathbf{0}$.
• If $a = \dfrac{1}{2}$, $A^k$ converges to $\dfrac{1}{2} \begin{pmatrix}1 & 1 \\\\ 1 & 1 \end{pmatrix}$.
• If $a = -\dfrac{1}{2}$, $A^k = (-1)^k \dfrac{1}{2} \begin{pmatrix}1 & 1 \\\\ 1 & 1 \end{pmatrix}$ and doesn’t converge.