Or are there examples where you can prove that the existing bijections are all uncomputable?

<Like I am not used to post... I will post little pieces of my work. If someone wanted to follow the whole process, just go to my account and look for my posts. The last ones would be the series of posts about my work.>

If we "could" build the next serie of "facts":

|A| not bigger than |B| not bigger than |C| not bigger than |D|

We could say: |A| is not bigger than |D|

A concrete Construction LJA, is an abstract data structure that helps us to create THAT serie of "facts" and intermediate "sets", in a particular example, between two different sets. Mostly between N being "D", and another set being "A". B uses to be a set of "representations", designed for THAT case. And "C" uses to be what we are going to call "Lists of finite paths" (LCF: 'Lista de Caminos Finitos' in spanish), designed for that concrete case too.

The data structure helps us, as a nemothecnic tool, to create the relations and the sets involved in them. The frustating stuff is once you have designed the relations, that helps you to prove in each step:

|I| is not biggert than |J|

You don't need the data structure anymore: evrything works fine... but seems very strange. But it works, if we take definitions as they are. We don't understand WHY the sets are the way they are, or why relations works so fine. The only way to understand all, is talking about the concrete example of CLJA we used to design each step and set, for that coparison of sets.

If we change the set "A"... we need to redesign everything. It would be another "concrete example", but the technic is always very very similar.

Like once a guy tried to publish a work with the same subject as mine... and he didn't publish formulas, let me publish in this post formulas, written in python, and some results, between the set LCF and N, for the comparison between P(N) and N.

PLEASE: realize THIS <is> a step of many... it is just "a piece" of all the work... and LCF is a set that you won't understand. Probably, the formula is impossible to understand just with the code. Let me put some links, and finally I will explain, a little, members of LCF.

This is the code in python of the bijection Omega: LCF -> N. It has the inverse function too... and it is not using prime numbers. In code, Omega Bijection is called "flja" (Function LJA).

https://github.com/CLJAs/clja-ftc

<I know I know.. it needs refactorization a lot.. but it works. I have many ideas but low energy>

T

It's a question way above my head, and I've never heard of there being a number system on a set with that cardinality.

It takes some work to go from the naturals to the reals via a power set operation. You can take the naturals including 0, {0,1,2,3,...} and then represent a choice of elements from that set as a binary sequence like 00111001... = {2,3,4,6,...} and so on. Each binary sequence can represent a real number if you work in a binary base system and let each term be a digit after the binary point. So zero is the empty set, one is the entire set of naturals, a half is the set {1}, and the binary number 0.0011101... is the set {2,3,4,6,...} and so on. There's a slight problem because in binary, for example, 0.00111111111... equals 0.01 (in the same way 0.999...=1 in decimal). I found this stackexchange post about that issue but I don't really understand the process:

https://math.stackexchange.com/questions/553526/the-set-of-real-numbers-and-power-set-of-the-natural-numbers

Once you fix that problem though, I believe you have bijection between the power set of the naturals and the some interval on the real line that you can put in bijection with the whole real line.

That does make me think though, if we can take the power set of the naturals and get a set that we can define operations on such as division, square roots, logarithms, etc, and we can call that set the real numbers, then if we take the power set of the reals, can we then define some operations on that new set and have some new number system? This seems to be a hard question to answer because we had to do some work with the power set of the naturals to get the reals.

I can't imagine how you would express an element of the power set of the reals, other than the clunky way of explicitly writing out something like {0,pi,sqrt(2)}, or {0,1,2,3,...} or {pi/2, 2pi/2, 3pi/2, 4pi/2,...}. You can't just make a binary sequence like 0011011... to represent a choice of the elements of the reals because there's no successor function any more. There's no next real number after 0. You need some way of indicating which elements you've chosen, but you can't do it in a list because the reals are unlistable.

It's fun to think about but I think the power set of the reals is too complicated to figure out any sort of straightforward and natural arithmetic on. The power set of the reals is sitting there though, probably with some operations and arithmetic that you could do with it that would be out of our g

Are there more integers than even numbers?

Common sense tells me there are obviously twice as many integers as there are even numbers. This is because the entire infinite set of integers should be comprised 50% of the infinite set of odd numbers, and 50% of the infinite set of even numbers.

But I'm trying to understand the concept of bijection as a method for comparing sizes of infinite sets, and it's been leading me to the opposite conclusion. As I understand it so far, we can determine that two infinite sets are of an equivalent size if we can set up a bijection between the two sets. In the example of integers vs even numbers, we should be able to set up a clear bijection by listing all even numbers sequentially (2, 4, 6, 8, 10...) and then assigning them a sequential integer based on where they fall in the sequence (1st even number, 2nd, 3rd, 4, 5...).

Where is my gap in comprehension? Am I right to think that the infinite set of integers is obviously twice as big as the infinite set of even numbers? Or am I right to think that, because there is a clear bijection between the two sets that makes the set of even numbers countably infinite, both infinite sets should be viewed as the same size?

Hi there!

So for my final in Discrete math, I have to present a proof. My probelm is as follows:

Let A, B, and C be sets. Let f: A to B and g: B to C. If f and g are both bijections, then g ⚬ f is a bijection.

The only issues I am having is creating a graph and knowing how to start it. Any help without solvin the problem for me is much appreciated! Thanks!!

Luckily, the next one was 121

Hello, currently I am working out of John Lee's Introduction to Topological Manifolds. I am learning about homeomorphisms right now, and the definition he gives is: a homeomorphism from X to Y is a bijective map f: X -> Y such that both f and f^(-1) are continuous.

This makes sense to me. I also understand that bijections automatically have a unique inverse. Next, he goes on to give an example of a homeomorphism from the unit sphere in R^(3) to the cubical surface of side 2 centered at the origin (denoted C). He gives the function f: C -> S^(2) and then as an exercise, gives the reader a function f^(-1) and says to "Show that f is a homeomorphism by showing that its inverse can be written as follows:" (and then writes the inverse).

Here's where the confusion starts - In the next paragraph, he basically says "Be careful, because if a function is bijective (i.e., has an inverse), that does not automatically mean its inverse is also continuous. This is fair to me. But I'm not understanding how we were able to "skip" this warning to prove the previous problem by showing f^(-1) was indeed the inverse to f (apparently implying that f is a homeomorphism if this is true).

Even still, he goes on in an exercise after that and asks the reader to prove that if f: X -> Y is a bijective continuous map, that the following are equivalent:

(a) f is a homeomorphism

(b) f is an open map

(c) f is a closed map

Which, part (a) I thought was quite actually the thing he just said wasn't always true? I feel like I am missing a huge piece of information or maybe misinterpreting something?

Anyone have any idea what I'm missing here? I feel like this is one of those things that is probably super obvious but I'm just completely reading over it somehow.

Definitions

Let a partition of N be [defined as](https://en.wikipedia.org/wiki/Partition_(number_theory)) (standard definition) a list of integers λ1, λ2, λ3, ..., λn such that λ1 + λ2 + λ3 + ... + λn = N and λ1 >= λ2 >= λ3 >= ... >= λn.

Let the Durfee rank of a partition be defined as (standard definition) the largest integer, s, such that the partition contains at least s parts with values ≥ s.

Let the Rakeof a partition, p, be defined as the minimum number of lists of parts of p such that for each list, L, all parts in L differ by no more than 1 (this is non-standard... see examples if this is confusing).

Now let D(n, k) be the set of all partitions of N = n and Durfee rank = k. And let R(n, k) be the set of all partitions of N = n and Rake = k

Examples

ptn = (5,4,3,2,1) | Durfee(ptn): 3 | Rake(ptn): 3

ptn = (5,5,5,5,5) | Durfee(ptn): 5 | Rake(ptn): 1

ptn = (11,7,6,4,2) | Durfee(ptn): 4 | Rake(ptn): 4

ptn = (2,2,2,1,1,1) | Durfee(ptn): 2 | Rake(ptn): 1

ptn = (4,4,4,3,3,3,2,2,2,1,1,1) | Durfee(ptn): 3 | Rake(ptn): 2

Question

D(n, k) and R(n, k) for all n and k <= sqrt(n) (a necessary restriction given the definitions above), are of the same cardinality. In fact, there always exists a bijection between these two sets. What is the bijection? FYI, This problem is way harder than it appears to be.

Hints

> > #1. The Rake of a partition can be computed greedily.

> > #2. Look at the Young Diagrams of the partitions. What are the necessary conditions for a partition to have Durfee rank of k or Rake of k?

> > #3. Working with partitions can be difficult and it's often useful to "represent" (wink, wink) them with more mathematically familiar objects. Partitions are in bijection with Dyck paths on an λ x λ grid where λ is the largest part of the partition or the number of parts in the partition -whichever is larger. Now you can work with binary string representations of partitions e.g. (4,3,3,1) -> 10110010. Further, to normalize the length of these strings, you can append infinite 0's and 1's to the ends of the string e.g.: 10110010 -> ...00000010110010111111.... Visually, this is just extending the λ x λ grid to an infinite grid. Try playing around with these strings and

Not sure how to do these, any help would be greatly appreciated!

Functions a, b, c, d : N × N → N are defined by the rules below. In each case decide whether the function is injective (one-to-one), surjective (onto), neither or both (bijective). Justify your answers.

a(x, y) = xy

b(x, y) = x + y

c(x, y) = (x + y)^2 + x

d(x, y) = (m − 1)^2 + m + x − y, where m = max(x, y).

For functions c and d it may help to draw up a 4 × 4 grid and write the image of (x, y) in the cell (x, y)

It's obvious that f(x) = 1/(1-x) - 1/x is at least a continuous bijection between (0,1)⋂Q and some dense subset of Q, but I haven't proven whether or not the function is surjective among all rationals. Is there a well-known function that satisfies the above properties?

If X and Y are topological spaces, f:X->Y is a continuous injection and g:X-Y is a continuous surjection, I've think I've shown that there isn't necessarily a continuous bijection. But my counterexample involved topologies with very few open sets -- in fact X and Y each had only one nontrivial proper open set.

I've been unable to come up with a counterexample involving nicer topologies. For instance if X and Y are both Hausdorff then does the existence of continuous injections and surjections from X to Y guarantee a continuous bijection? I'm still guessing no, but haven't been able to construct a counterexample.

Any hints or suggestions would be welcome.

So, for it to be an isomorphism, sets X and Y must be the same size. Are not all sets Sx and Sy anyway isomorphic if X and Y are the same size?

The composition from right to left yields for an element of Y, an element of X (f^-1) then this is multiplied by α, presumably an element of X, and then mapped back to Y with f

As the function f is a bijection, and f^-1 is a bijection, α * x yields a permutation of X. Does this show the isomorphism?

Wikipedia is a bit overkill

Hi! Can someone explain why the implication if aH = bH then Ha^{-1} = Hb^{-1} proves that there is a bijection between left and right cosets?

Let R be a set of real numbers.

Does there exist a bijection between R and all subsets of R? My guess would be that there isn't but I am not sure how to approach it.

And a follow up question, is there a bijection between R and all finite subsets of R?

I thought that we could do this by stating that every unique box will have a some number of balls from 0 to n, and every ball will be placed in a uniquely labeled box. So they’re both invective on each other, and hence, bijective. But then my domain and co-domain do not match to the question’s, so I am gravely mistaken. The domain is the permutation set and the co-domain is the combination set. How do I prove this?

Is it possible to associate each real function with a real number so that each function has exactly one number associated with it and each real number has one corresponding function?

Or at least real functions that continious and differentiable at every point?

I have a doubt regarding this question.

>Let E , F be sets.
>
>f maps the set E to F.
>
>Show that for every u,v ∈ F if u ≠ v then f ^(-1) ({u}) ∩ f^(-1) ({v}) = ∅

If f was a bijection I could say that every pre-image is unique and since u ≠ v their preimages are unique.

Hence the intersection will be empty.

But here I make an assumption that f is a bijection. I understand that for a inverse function to exist the function has to be bijective.

I just want to know if my answer is correct?

Appreciate your time. Thanks in advance!

I've it's solution but I didn't get that.

SOLUTION ON THE BOOK

f (x) = 2x + sin x

f `(x) = 2 + cos x

here, f `(x) > 0 always (I didn't get this step)

Hence, fxn is one to one fxn, also range of f(x) is R and it's codomain is R => this fxn is onto too.

=> fxn is bijective

Let X,Y be Banach spaces and S,T : X -> Y two bounded linear operators. Show that if T is a bijection, there exists δ>0 such that if ||S-T||<δ then S is a bijection.

I don't have enough knowledge to solve this question, I just saw it on an old qualifying exam and it made me curious.

Hi, I am playing around with bijections and I wanted to prove that the cardinality of real numbers in [0,1] is the same as the cardinality of real numbers in (0,1] but I can’t think of any function. How can I do this?