real analysis - Finding a bijective Map

I need to find a bijective map from $A=[0,1)$ to $B=(0,1).$ Is there a standard method for coming up with such a function, or does one just try different functions until one fits the requirements?

I've considered some "variation" of the floor function, but not sure if $\left \lfloor{x}\right \rfloor$ is bijective. Now i'm thinking of maybe some trig function that takes elements of $A$ and maps them to $B$.

I know how to show that a function is injective and surjective, but how do I find such a function? Am I overthinking this or am I under thinking this?

Any help would be greatly appreciated!!!

Lastly, since we are to find a bijective map from $A$ to $B$, assuming one exists, is this equivalent to saying that these two sets have the same cardinality?

Thank you

Answer

Consider the function
$\psi: A \rightarrow B$ given by
$$
\psi(x) = \begin{cases} x & \text{ if}\; \; \; x \neq 0 \; \; \text{and}\; \; x \neq \frac{1}{n} \in \mathbb{N} \\
\frac{1}{2} & \text{if} \;\; x = 0 \\

\frac{1}{n+1} & \text{if } \; \; x= \frac{1}{n}, \; n \in \mathbb{N}
\end{cases}
$$
It is not difficult to verify that $\psi$ is bijective. The problem with constructing a bijection using our favorite elementary functions is that no continuous bijection between the sets exists. This is due to the differences in character (or topology) of the two sets. So unfortunately one must turn their attention to different functions.

The function above relies on the fact that we can get this bijection by taking a sequence in $(0,1)$, say $\{x_n\}$ and mapping $x_1$ to the extra point $0$, and then mapping $x_n$ to $x_{n-1}$ for all other $n$. Hopefully this trick helps with your future work.

real analysis - Continuous function from $(0,1)$ onto $[0,1]$

While revising, I came across this question(s):

A) Is there a continuous function from $(0,1)$ onto $[0,1]$?

B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?

(clarification: one-to-one is taken as a synonym for injective)

I figured the answer to A is yes, with $\frac{1}{2}\sin(4\pi x)+\frac{1}{2}$ as an example.

The answer to part B is no, but what is the reason?

Sincere thanks for any help.

Answer

There isn't such a function. If $f$ is such a function, then $f$ is monotonic, and $f^{-1}$ too. But a monotonic function defined on an interval is continuous iff its range is an interval. So we have that $f$ is an homeomorphism from $(0,1)$ to $[0,1]$, which is impossible since one is compact and the other is non compact.

Alternating series: $sumlimits_{n= 1}^{infty} (-1)^{n-1} frac{ln(n)}{n}$ convergence?

Determine whether the series converges absolutely, conditionally or diverges?

$$\sum\limits_{n= 1}^{\infty} (-1)^{n-1} \frac{\ln(n)}{n}$$

I know that $\sum|a_{n}|$ diverges by using the comparison test:

$$\frac{\ln(n)}{n} > \frac{1}{n}$$ and the smaller, r.h.s being the divergent harmonic series.

So, should my conclusion for the alternating series be divergent or convergent conditionally^*?

---

^{* How to estimate whether the alternating series terms are cancelling?}

Answer

Let $f(x)=\frac{\ln x}{x}$ so $f'(x)=\frac{1-\ln x}{x^2}\le 0$ for $x\ge e$ and so the sequence $\left(\frac{\ln n}{n}\right)_{n\ge3}$ is decreasing to $0$. Apply now the alternating series criteria to conclude the convergence of the series.

calculus - Why does the antiderivative of $frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$)

Why does $\log{x}$ represent the area below the graph of $f(x)=\frac{1}{x}$ from $1$ to $x$? What's so special about $1$ in this case?

Of course I understand that $\log{1}=0$ and I also understand that you cannot start at $x=0$ because $f(0)$ is not defined.

Still I can't get my head around of why it has to be $1$.

Also, this further implies that part of the antiderivative of $f(x)=\frac{1}{x}$ has to be negative (as part of the function of $\log{x}$ is negative). But why is this necessary?

The background of this question is that my calculus-book (Calculus, a complete course) starts with noticing that $f(x)=\frac{1}{x}$ is not an antiderivative of a polynomial function and then attempts to define a antiderivative which ends up being $\log{x}$. It does this all before even addressing the fundamental theorem of calculus or techniques of integration. They then simply define $log(1)$ to be $0$ without even knowing what that function really is yet. So I am stuck with kind of a circular reasoning where log(1)=0 because we defined it that way, but I don't understand why we define it that way.

When calculating areas under graphs by taking the limit of a sum (instead of by integration), you would start at $x=0$ right?

So in short: Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$). Why do we define it that way?

I am looking for some kind of deeper understanding; something still didn't "click". So some real background on what's going on here would be much appreciated.

Thanks!

Answer

What is "the" anti-derivative of $x\mapsto x+3$? Well, there isn't really a "the" here. For any constant $C$, the map $x\mapsto\frac12x^2+3x+C$ ia an anti-derivative. It looks like $x\mapsto \frac12x^2+3x$, obtained by picking $C=0$ plays a special role among the anti-derivatives, but it doesn't. Somebody else might rightfully describe the family of anti-derivatives of $x\mapsto x+3$ as $x\mapsto \frac12(x+3)^2+C$ (and in fact this can happen in a systematic way by using the substitution rule when integrating), again with $C$ an arbitrary constant, but unrelated to the $C$ used above. Of course, if we set $C=0$ in this description, we arrive at $x\mapsto (x+3)^2$, not the same function as $x\mapsto \frac12x^2+3x$. So is one of them "better" than the other? This may be the case depending on context (e.g., the second form tells us immediately where the apex of the parabola is), but there is no general notion of better.

So what is an anti-derivative of $f\colon(0,\infty)\to \Bbb R$, $x\mapsto \frac1x$? It is any function $F\colon(0,\infty)\to \Bbb R$ with $F'(x)=f(x)$ for all $x>0$. If we do not have a suitable function with this property already in our utility belt, we might simply define a specific one this way and give it a name. To get rid of the $C$, we rely on the definite integral, i.e., we fix some $a$ and define $F(x)=\int_a^xf(t)\,\mathrm dt$. This merely shifts the burden: What $a$ do we pick?
$a=0$ would be nice, but $0$ is not in the domain of $f$; and not even $\lim_{a\to0}\int_a^xf(t)\,\mathrm dt$ exists. We might even try $a=+\infty$, but alas, $\lim_{a\to+\infty }\int_a^xf(t)\,\mathrm dt$ doesn't exist either (but for other $f$ this might still be a viable choice; for example, in physics, this is the default way to pick the integration constant when defining potentials). So we better pick $a$ somewhere between $0$ and $\infty$. But as $f(x)>0$ for all $x$, this automatically implies $F(x)a$, i.e., we cannot achieve that $F$ is non-negative (or non-positive) throughout.

But does that mean $a=1$ is the best choice? Maybe $a=\pi$ woul dmake things particularly elegant?
Well, as above, this may depend on the context, but there are some overwhelming contexts:
By substituting $x\leftarrow \frac1cx$ (with $c>0$), we see that
$$F(b)=\int_{a}^bf(x)\,\mathrm dx=\frac1c\int_{ac}^{bc}f({\tfrac xc})\,\mathrm dx=\frac1c\int_{ac}^{bc}\frac1c{x}\,\mathrm dx=\int_{ac}^{bc}\frac1{x}\,\mathrm dx=F(bc)-F(ac)$$
or: $F(bc)=F(b)+F(ac)$. This formula becomes most beautiful (namely: symmetric) if we agree on $a=1$: $F(bc)=F(b)+F(c)$.
This is one compelling reason to define our "go-to" anti-derivative (which we decide to call $\ln$) as $\ln x:=\int_1^x\frac 1t\,\mathrm dt$.

real analysis - Compute $sumlimits_{n=1}^inftyfrac{1}{(n(n+1))^p}$ where $pgeq 1$

I was recently told to compute some integral, and the result turned out to be a scalar multiple of the series $$\sum\limits_{n=1}^\infty\frac{1}{(n(n+1))^p},$$ where $p\geq 1$. I know it converges by comparison for
$$\dfrac{1}{(n(n+1))^p}\leq\dfrac{1}{n(n+1)}<\dfrac{1}{n^2},$$
and we know thanks to Euler that $$\sum\limits_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.$$ I managed to work out the cases where $p=1$ and $p=2$. With $p=1$ being a telescoping sum, and my solution for $p=2$ being $$\frac13\pi^2-3,$$ which I obtained based on Euler's solution to the Basel Problem. I see no way to generalize the results to values to arbitrary values of $p$ however. Any advice on where to start would be much appreciated.

Also, in absence of another formula, is the series itself a valid answer? Given that it converges of course.

combinatorics - Prove König's theorem using Dilworth's theorem

I am trying to derive König's theorem from Dilworth's theorem, but it seems like I'm stuck. I know that I have to define some kind of binary relation on the set of a bipartite graph's vertices, then use Dilworth's theorem. However, every relation I've come up with so far either did not satisfy some properties needed for Dilworth's theorem or seemed to be completely useless. Can anyone point me in the right direction?

Thanks.

abstract algebra - Linearly disjoint?

Let $f$ and $g$ be irreducible polynomials over a field $K$ with $\deg f=\deg g =3$ and let the discriminant of $f$ be positive and the discriminant of $g$ negative.

Does it follow that the splitting fields of $f$ and $g$ are linearly disjoint? If yes, why?

(Def.: Two intermediate fields $M_1, M_2$ of an algebraic field extension $L|K$ are called linearly disjoint, if every set of elements of $M_1$, that is linearly independent over $K$, is also linearly independent over $M_2$.
Here, this is equivalent to $M_1 \cap M_2 = K.$)