real analysis - Proving part of the theorem that a continuous function on the bounded interval is uniformly continuous

I am asked to prove the following. Suppose $D$ is a closed and bounded subset of $R$ and suppose $f:D\to R$ is continuous on $D$. Then $f$ is uniformly continuous:
$

$Proof: Suppose by contradiction that $\{x_n\}$ and $\{y_n\}$ are sequences in $D$ s.t.
$$ \lim_{n\to\infty} [x_n-y_n] =0$$

and the sequence $[f(x_n)-f(y_n)]$ does not converge to $0$.

We want to negate the statement of convergence for the above s.t. we have: "There exists an $\epsilon>0$ s.t. for all natural number k there exists an $n_k>k$ s.t. |$f(x_{n_k})-f(v_{n_k})$|$\geq \epsilon$.

Now apply the sequential compactness theorem to $\{x_{n_k}\}$ which is a subset of $D$. There is a limit point $x_0$ s.t. there is a further subsequence $\{x_{n_{k_j}}\}$ s.t. the limit as j tends to infinity is $x_0$, which is in $D$.

Now here's what I am supposed to prove: By supposing that $\{x_n\}$ and $\{y_n\}$ are sequences satisfying the fact that the image of a continuous function on a closed bounded interval is bounded above and assuming $\{x_n\}$ converges to $x$ I want to show that the limit as $j$ tends to infinity of $y_{n_{k_j}}$ is $x_0$. How would you do this part?

So basically I am just proving this one last part of the proof.

complex analysis - Where does this equation come from?

Since I study 3 years i ask myself very often where does this equation come from?

$$e^{i\theta} = \cos(\theta)+i \sin(\theta)$$
Is it found by series expansion?

Answer

This result is commonly shown via Taylor series, as explained in the comments, and is well-known. I'd like to offer a different sort of proof, for those who are interested, that I believe is easier yet less well-known.

Consider the second order linear differential equation
$$y''=-y$$
We know the most general solution is:
$$y = A\cos{x}+B\sin{x}$$
But $$y = e^{ix}$$ is also a solution, and by existence and uniqueness theorems, that means $$e^{ix} = A\cos{x}+B\sin{x}$$

for some $A,B$. Plugging in $x=0$ for the expression and its first derivative, we see that $A = 1, B = i$.

Thus, $$e^{ix} = \cos{x}+i\sin{x}$$

elementary set theory - Is the class of subsets of integers countably infinite?

I am trying to figure out if the class of subsets of integers is countably infinite or not. I know that a collection of all finite subsets of integers is countable. I believe i need to use diagonalization to prove whether it's true or not but I'm not sure how to approach it.

All Help is greatly appreciated!

Answer

No, the set of all subsets of the integer is not countable. Since $\mathbb{Z}$ has the same cardinality as $\mathbb{N}$, it suffice to consider all subsets of $\mathbb{N}$.

For each subset $X$ of $\mathbb{N}$ consider the characteristic function $\chi_X$ defined by

$\chi_X(z) = \begin{cases}

1 & \quad z \in X \\
0 & \quad z \notin X
\end{cases}$

In this way you associate injectively and subjectively each subset $X$ of $\mathbb{N}$ with a function in $2^{\mathbb{N}}$. $2^\mathbb{N}$ has cardinality strictly larger than $\mathbb{N}$. This is proved by the typical Cantor diagonalization argument.

Also, Cantor Diagonalization and the function I wrote above can be used to show more generally that the set of all subsets of a given set has cardinality strictly greater than the given set.

In response to comment :

You can think of a function from $\mathbb{N} \rightarrow 2$ a infinite binary strings of $0$'s and $1$'s. Assume that $2^{\mathbb{N}}$ is countable. That is there is a bijection $\sigma$ from $\mathbb{N}$ to $2^\mathbb{N}$. Then define the function $h : \mathbb{N} \rightarrow 2$ as follows

$h(n) = \begin{cases}
1 & \quad (\sigma(n))(n) = 0 \\
0 & \quad (\sigma(n))(n) = 1
\end{cases}$

Informally, this is the familiar argument, form a new binary string by going down the diagonal and switching $0$ for $1$ and $1$ for $0$. Now this is a a perfectly good binary string hence it down appear as $\sigma(k)$ for some $k$ if $\sigma$ is indeed a bijection. However, it can not be $\sigma(k)$ for any $k$ since it differs from $\sigma(k)$ in at least the $k^\text{th}$ entry.

I hope this helps.

calculus - Prove $sum_{n geq 0}c_n$ converges iff $sum_{k geq 0}(c_{2k}+c_{2k+1})$ converges

Suppose the sequence $\{c_n\}$ converges to zero. Prove $\sum_{n \geq 0}c_n$ converges iff $\sum_{k \geq 0}(c_{2k}+c_{2k+1})$ converges. Moreover, if the two series converge then they have the same limit.

I was thinking that if $$\lim_{n \rightarrow \infty}c_n \neq 0$$
Then the statement can't be true. Like the sequence
$$1, -1, 1 ,-1 , 1 ,-1, \cdots$$

So $$\sum_{k \geq 0}(c_{2k}+c_{2k+1})=0$$
which is convergent, but
$$\sum_{n \geq 0}c_n \text{does not exists}$$
So how could we prove under the assumption $\lim_{n \rightarrow \infty}c_n=0$? Many thanks~

Answer

If $\sum_{n=0}^{\infty}c_n = a$ then we have $$S_n = \sum_{k=0}^n c_k \to a.$$ So, $$ \sum_{k=0}^n (c_{2k} + c_{2k+1}) = S_{2n+1} \to a. $$

I'll let you handle the (slightly messier) other direction.

real analysis - Can we construct a continuous but nowhere differentiable surface?

In my analysis course we learned about the Weierstrass function which is continuous but nowhere differentiable, is it possible to make a surface which is continuous and nowhere differentiable?

Answer

If $W$ is Weierstrass function, let $z=W(x)\,W(y)$.

real analysis - How to find $lim_{ntoinfty}int_{1/n}^{2/n}frac{f(x)}{x},mathrm{d}x$?

Suppose $f:\mathbf{R}\to\mathbf{R}$ is continuous and $f(0)=3$. Find $$\lim_{n\to\infty}\int_{1/n}^{2/n}\frac{f(x)}{x}\mathrm{d}x.$$

I don't know how to do this problem, any hints?

Answer

$f$ is continuous at $0$, hence there exists $\delta>0$ such that $|f(x)-3|<\delta$ if $|x|<\varepsilon$. It follows that the integral $L$ is such that
$$
(3-\delta)\log 2=\int_{1/n}^{2/n}\frac{3-\delta}{x}\mathrm{d}x\le L\le \int_{1/n}^{2/n}\frac{3+\delta}{x}\mathrm{d}x=(3+\delta)\log 2
$$
if $n$ is sufficiently large. Hence the limit is $3\log 2$.

(Thanks to Michael Hardy and Winther for spotting a mistake.)

real analysis - Determine if $lim_{nrightarrow infty }int_{-infty}^{infty}frac{(sin(x))^{n}}{x^2+1}dlambda(x)$ exists and calculate its value if it exists.

Determine if the next limit exists and calculate its value if it exists.
$$\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}\frac{(\sin(x))^{n}}{x^2+1}d\lambda(x)$$
where $\lambda$ is the Lebesgue measure in $\mathbb{R}$

My attepmt: We consider $f_{n}(x)=\frac{(\sin(x))^{n}}{x^2+1}$, and $g(x)=\frac{1}{x^2+1}$. Note that $g$ is absoltely improper Riemann integragle, that is, $\int_{-\infty}^{\infty}|g(x)|dx<\infty$ because
$$\int_{-\infty}^{\infty}|g(x)|dx=\int_{-\infty}^{\infty}g(x)dx=\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx=2\int_{0}^{\infty}\frac{1}{x^2+1}dx=\pi<\infty.$$
Therefore, $g$ is Lebegsue integrable in $[-\infty,\infty]$ and
$$\int_{-\infty}^{\infty}g(x)d\lambda(x)=\int_{-\infty}^{\infty}g(x)dx=\pi <\infty.$$

Note that $|f_{n}(x)|\leq g(x)$ for all $x\in [-\infty,\infty]$. Also, $\{f_{n}(x)\}_{n}$ converge for each $x\in [-\infty,\infty]$, then we define $f$ as
$$f(x)=\lim_{n\rightarrow \infty }f_{n}(x)=\left\{\begin{array}{ll}1 & \mathrm{if}\:x=\frac{\pi}{2}+2m\pi \mbox{for some }m\in\mathbb{N} \\
-1 & \mathrm{if}\:x=\frac{3\pi}{2}+2m\pi \mbox{ for some }m\in\mathbb{N}\\ 0 & \mbox{otherwise} \end{array}\right.$$

We have that $\int_{-\infty}^{\infty}f d\lambda(x)=0$ because $f$ is not zero in a discrete set. Is it correct?

Then, by Dominated Convergence Theorem we have that $f$ is Lebesgue integrable and

$$\begin{array}{rcl}\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}\frac{(\sin(x))^{n}}{x^2+1}d\lambda(x)&=&\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}f_{n}(x)d\lambda(x)\\ &=&\int_{-\infty}^{\infty}\lim_{n\rightarrow \infty }f_{n}(x)d\lambda(x) \\ &=&\int_{-\infty}^{\infty}f(x) d\lambda(x)\\&=&0.\end{array}$$

The question: Is my attempt correct? Do I have an error? and if I have it, what is it?

Answer

The only small mistake is that for $x$ of the form $3\pi/2+2\pi m$, $f_n(x)$ is of the form $(-1)^n c$ where $c$ do not depend on $n$ hence the sequence $\left(f_n(x)\right)_{n\geqslant 1}$ is not convergent.

However, the almost everywhere convergence is sufficient to apply the dominated convergence theorem. Actually, we do not even need to see what happens on the set $N:=\left\{\pi/2+2\pi m,m\in\mathbb Z\right\}\cup \left\{3\pi/2+2\pi m,m\in\mathbb Z\right\}$ since its measure is zero.