measure theory - Do limits of sequences of sets come from a topology?

In measure theory we frequently see the following definitions:

$$\limsup_{n\to\infty} A_n = \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} A_j\right)$$

$$\liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} A_j\right)$$

where $(A_n)_n$ is a sequence of measureable sets i.e. $\forall n: A_n\in\mathcal{M}$, where $\mathcal{M}$ is a $\sigma$-algebra on $X$, for example $\mathcal{M} = 2^X$. Therefore it makes sense to also define:

$$\lim_{n\to\infty}A_n = \limsup_{n\to\infty} A_n = \liminf_{n\to\infty} A_n$$

when the last two agree. If $\mu$ is a finite (positive, to keep things simple) measure, it is easy to see that under such definition we have $\mu(\lim_{n\to\infty}A_n) = \lim_{n\to\infty}\mu(A_n)$, whenever $\lim_{n\to\infty}A_n$ exists, which looks like some kind of continuity.

Does this kind of convergence of sequences of measurable sets arise from a (preferably Hausdorff, so that limits are unique) topology on $\mathcal{M}$? If such a topology exists, is $\mu:\mathcal{M}\to[0,\infty)$ in fact a continuous function?

(A related question that may be of interest would be: what happens if we allow arbitrary sets? Can we make the Von Neumann universe $V$ into a topological space in such a way?)

Answer

There is such a topology. Simply give $\mathcal{M}$ the subspace topology induced by the product topology on $2^X$.

It may help to think of $2^X$ as the set of functions from $X$ to $\{0,1\}$, by identifying a set with its indicator function. Then we have $1_{\limsup A_n} = \limsup 1_{A_n}$ and so on. Since the product topology is just the topology of pointwise convergence, this behaves as desired.

However, the map $A \mapsto \mu(A)$ is not in general continuous with respect to this topology. For instance, the finite sets are dense in $\mathcal{M}$ with this topology, and so any nontrivial non-atomic measure gives a discontinuous map.

limits - Is there a standard way to compute $limlimits_{ntoinfty}(frac{n!}{n^n})^{1/n}$?

I'm computing the radii of convergence for some complex power series. For one I need to compute
$$\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}.$$

I know the answer is $\frac{1}{e}$, so the radius is $e$. But how could you compute this by hand? I tried taking the logarithms and raising $e$ by this logarithm, but it didn't lead me to the correct limit. (This is just practice, not homework.)

Answer

There are two formulas to compute radius of convergence of the series $\sum\limits_{n=1}^\infty{c_n}z^n$
$$
\frac{1}{R}=\lim\limits_{n\to\infty}|c_n|^{1/n}=\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|.
$$
Use the second one.

combinatorics - How many strings contain every letter of the alphabet?

Given an alphabet of size $n$, how many strings of length $c$ contain every single letter of the alphabet at least once?

I first attempted to use a recurrence relation to work it out:

$$
T(c) = \left\{ \begin{array}{cr}
0 &\mbox{ if $cn! &\mbox{ if $c = n$} \\
T(c-1) \cdot n \cdot c &\mbox{ if $c > n$}
\end{array} \right.
$$

As there's no strings that contain every letter if c < n, and if c = n then it's just all permutations. When c > n you can take any string of size (c-1) that contains all letters (of which there are $T(c-1)$ to choose from), you choose which letter to add (of which there are $n$ choices) and there are $c$ different positions to put it. However, this gives out results that are larger than $n^c$ (the total number of strings), so it can't be right, and I realised it was because you could count some strings multiple times, as you can make them taking different inserting steps.

Then I thought about being simpler: you choose n positions in the string, put each letter of the alphabet in one of those positions, then let the rest of the string be anything:

$$
{c\choose{n}} \cdot n! \cdot n^{c-n}
$$

But again this counts strings multiple times.

I've also considered using multinomial coefficients, but as we don't know how many times each letter appears in the string it seems unlikely they would be much help. I've also tried several other methods, some complicated and some simple, but none of them seem to work.

How would you go about working out a formula for this? I'm sure there's something simple that I'm missing.

Answer

Let $W(c,n)$ denote the number of words of length $c$ from an alphabet of $n$ letters. Then $W(c,n)=n^c$.

Out of these, the number of words of the same size that do not contain one of the letters is $W(c,n-1)=(n-1)^c$. The number of ways of choosing which letter is missing is $\binom{n}{1}$.

The number of words of the same size that do not contain two letters is $W(c,n-2)=(n-2)^c$. The number of ways of choosing which two letters are missing is $\binom{n}{2}$... and so on ...

Now we use inclusion-exclusion principle: (subtract the number of words missing one of the letters, then add the number missing two of the letters, add the number missing three of the letters,...)

We get:

$$W(c,n)-\binom{n}{1}W(c,n-1)+\binom{n}{2}W(c,n-2)-\binom{n}{3}W(c,n-3)+...+(-1)^{n-1}\binom{n}{n-1}W(c,n-(n-1)).$$

This is

$$n^c-\binom{n}{1}(n-1)^c+\binom{n}{2}(n-2)^c+\binom{n}{3}(n-3)^c+...+(-1)^{n-1}\binom{n}{n-1}1^c.$$

or

$$\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}(n-k)^c.$$

Another way could be: Denote $S_c^n$ the number of ways to partition the word of length $c$ into $n$ pieces. Then we just need to choose which letter goes to each of the $n$ pieces. This number is $n!$. So the number of words we are looking for is

$$n!S_c^n.$$

The numbers $S_c^n$ are called Stirling's numbers of the second kind.

real analysis - Showing $ sum_{n=0}^{infty} frac{1}{(3n+1)(3n+2)}=frac{pi}{3sqrt{3}}$

I would like to show that:

$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$

We have:

$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$

I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...

Answer

Regularized the series:
$$ \begin{eqnarray}
\sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} &=& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) = \sum_{n=0}^m \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
&=& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x =
\int_0^1 \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x
\end{eqnarray}
$$

Now we can take the limit by dominating convergence theorem:
$$
\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2 \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}}
$$

The sum of the cubes of two different prime numbers is 5256.

Is an simple way to solve the problem?

The sum of the cubes of two different prime numbers is 5256. What are the two primes?

Here is what I did: assume the two numbers are $x$ and $y$, then I have $x^3+y^3=5256$. I can try x and y, but I don't think it is the most effective method. Can anyone help me?

linear algebra - A real function which is additive but not homogenous

From the theory of linear mappings, we know linear maps over a vector space satisfy two properties:

Additivity: $$f(v+w)=f(v)+f(w)$$

Homogeneity: $$f(\alpha v)=\alpha f(v)$$

which $\alpha\in \mathbb{F}$ is a scalar in the field which the vector space is defined on, and neither of these conditions implies the other one. If $f$ is defined over the complex numbers, $f:\mathbb{C}\longrightarrow \mathbb{C}$, then finding a mapping which is additive but not homogenous is simple; for example, $f(c)=c^*$. But can any one present an example on the reals, $f:\mathbb{R}\longrightarrow \mathbb{R}$, which is additive but not homogenous?

Answer

If $f : \Bbb{R} \to \Bbb{R}$ is additive, then you can show that $f(\alpha v) = \alpha f(v)$ for any $\alpha \in \Bbb{Q}$ (so $f$ is a linear transformation when $\Bbb{R}$ is viewed as a vector space over $\Bbb{Q}$). As $\Bbb{Q}$ is dense in $\Bbb{R}$, it follows that an additive function that is not homogeneous must be discontinuous. To construct non-trivial discontinuous functions on $\Bbb{R}$ with nice algebraic properties, you usually need to resort to the existence of a basis for $\Bbb{R}$ viewed as a vector space over $\Bbb{Q}$. Such a basis is called a Hamel basis. Given a Hamel basis $B = \{x_i \mid i \in I\}$ for $\Bbb{R}$ (where $I$ is some necessarily uncountable index set), you can easily define a function that is additive but not homogeneous, e.g., pick a basis element $x_i$ and define $f$ such that $f(x_i) = 1$ and $f(x_j) = 0$ for $j \neq i$.

trigonometry - $sinalpha + sinbeta + singamma = 4cos{frac{alpha}{2}}cos{frac{beta}{2}}cos{frac{gamma}{2}}$ when $alpha + beta + gamma = pi$

Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle)

Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

---

There is already a solution on Math-SE, however I want to avoid using the sum-to-product identity because technically the book I go by hasn't covered it yet.

So, is there a way to prove it with identities only as advanced as $\sin\frac{\alpha}{2}$?

---

Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).

---

Answer

You may go the other way around:
$$
\cos\frac{\gamma}{2}=\cos\frac{\pi-\alpha-\beta}{2}=
\sin\frac{\alpha+\beta}{2}=
\sin\frac{\alpha}{2}\cos\frac{\beta}{2}+
\cos\frac{\alpha}{2}\sin\frac{\beta}{2}
$$
so the right hand side becomes
$$

4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}
\sin\frac{\alpha}{2}\cos\frac{\beta}{2}+
4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}
\cos\frac{\alpha}{2}\sin\frac{\beta}{2}
$$
Recalling the duplication formula for the sine we get
$$
2\sin\alpha\cos^2\frac{\beta}{2}+2\sin\beta\cos^2\frac{\alpha}{2}
$$
and we can recall

$$
2\cos^2\frac{\delta}{2}=1+\cos\delta
$$
to get
$$
\sin\alpha+\sin\alpha\cos\beta+\sin\beta+\sin\beta\cos\alpha
=
\sin\alpha+\sin\beta+\sin(\alpha+\beta)=
\sin\alpha+\sin\beta+\sin\gamma
$$