Uncountable set Puns

How do you prove that the set of all non decreasing functions from N -> N is uncountable

Let N = {1, 2, 3, 4, 5, ...} and let S be any set. (What exactly is the definition of natural numbers to follow when talking about countable sets? Cause some definitions include 0 while others do not. I'm under the impression that {1, 2, 3, 4, 5, ...} is more proper.)

Some materials online state that:

Def. 1: "A set S is said to be countable if there exists a bijection between N to S, i.e. f : N → S."

While others state that:

Def. 2: "A set S is said to be countable if there exists a bijection between S to N, i.e. f : S → N."

I know this might not be that big of a deal since bijective functions are known to be invertible but it makes me wonder, especially with respect to proofs involving countability, on what definition is better to follow.

Ex. Show that the set of integers, Z, is countable.

Proof using Def. 1 (more common from my observation):

Let f : N → Z be defined as:

f(x) = x/2 if x is even and f(x) = -(x-1)/2 if x is odd. Proof by cases since piecewise, show bijection.

Ends up showing that N = {1, 2, 3, 4, 5, ...} maps 1-1 to Z = {0, 1, -1, 2, -2, ...}.

Proof using Def. 2:

Let f : Z → N be defined as:

f(x) = 2x if x>0 and f(x) = -2n + 1 if x is <= 0. Proof by cases since piecewise, show bijection.

Ends up showing that Z = {0, 1, -1, 2, -2, ...} maps 1-1 to N = {1, 2, 3, 4, 5, ...}.

Thank you so much.

I'm having trouble getting started.

How do I show that if we have an uncountable subset S in [0,1]x[0,1] in R^2, then there is a point p in S such that for any e>0, the closed ball B(p,e) contains uncountably many points in S?

I don't even know where to begin.

I would prefer a simple explanation since I am struggling with this.

The argument I’ve seen is just that it isn’t surjective since a A set that holds values that N doesn’t map to in a map from N to the P(N). Then if A is the same map there is a contradiction because anything plugged into the map can’t be in A. The part that makes no sense is that if there is this set A then of course it isn’t surjective.

If N mapped to N somehow had this set A then it wouldn’t be surjective by this logic since the argument doesn’t show that the set would be empty.

Since in the case of N to N it would have to be empty for N to be surjective

I’m confused because N, N^2, and N∗ are all countable.

To be more specific my question ask me to prove the following:

Say there is a "double n" is a set of unordered pairs with {0,1,2,3...n} (ex: double 3 would be a set with 10 elements (0,0),(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)). And say there is a chain such that the second value of the unordered pair is the same as the first value of the next pair (ex: (0,1),(1,2),(2,2),(2,0),(0,3))

I need to prove that there is an uncountable number of infinite-length chains that can be formed using "double N", where N is the set of all natural numbers.

My though process: I think I should assume we have a bijection between N and double N. And then construct an element from double N that is not in the image of the bijection function. I tried applying Cantor's proof on powerset where he used the diagonal thing but I can't seem to find a way to make it work.

Thanks in advance

I know that the question seems obvious, but I'm still struggling to show it in a formal way. If a set A is uncountable, then there is no injective map f: A -> N, where N is the set of natural numbers. But how do we argue (w/o waving hands) that A has a countable subset?

Edit: New version. Made the reasoning at the end more clear.

I am wondering what anyone thinks of this write-up.

The idea is supposed to be simple, but it took me a few pages to explain it. The actual proof is just the last paragraph on the last page.

I there is mistake or suggestion for improvement, I would really want to hear about it.

Has this line of reasoning been discovered previously? If so, I'd love to see a reference.

As the set of statements in a formal system can be at most countable infinite - if I understand it correctly - can e.g. mathematical statements over R or just the set of all real numbers be a formal system?

If they are not formal systems, what are they then? Or is it possible to just circumvent this problem by definind an uncountable set of variables or symbols?

Many thanks in advance.

For all uncountable sets X, does there exist a set Y such that |X| = |P(Y)|?

(Where P denotes the powerset.)

Just re-reading my old set theory notes out of boredom.

I've noticed that in proving that an uncountable ordinal exists, there is no requirement to have the powerset axiom. This surprises me somewhat, since looking at the axioms on there own the powerset one looks like the only one that can take you beyond the countable world. I'm struggling to see which other axiom you can use to create an uncountable set. Replacement preserves cardinality, and if you only have countable sets then unions also preserve cardianlity (this may need choice).

It's well known that without AoI all sets could be finite, but with AoI, which only states that a countable set exists, you suddenly get the whole ordinal heirachy.

Is there then a simplified proof of the existence of an uncountable set from ZF - PS, which highlights where exactly the gap between countable and uncountable is bridged?

I've been working on this single problem on my homework for hours (mainly referring to the book and looking up similar solutions), but I just do not get this question:

Show that the set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is uncountable.

Could someone walk me through this logic? It would be much appreciated.

I'll use real numbers in examples as it is the uncountable set I'm most familiar with, if a different set makes a better example feel free to introduce me.

To the extent that it is impossible to write an algorithm to touch "each" entry in an uncountable set does it even have meaning to try to describe such an operation.

For example, I feel like I should be able to trivially say that the product of all real numbers is 0. 0 is a real number; multiplication in the reals commutative; multiplication in the reals is closed; and 0 times any real number is 0. So whatever the product of all real numbers except 0 is, if you multiply it by 0 you get 0. Is that rigorous enough to hand wave away "whatever the product of all real numbers except 0 is" when I know it is not only not computable, but not even mathematically describable?

I feel like I should be able to say the sum of all real numbers is 0. Since multiplication is closed over the reals, for every real number K there exists a different real number -1 * K except when K =0; K + -1 * K = 0; for any real number C, 0 + C = C. So 0 + however you decide to add up all the other 0's is still going to give you 0. This "process" is even more hand-wavy though as it assumes computing and summing an uncountably infinite number of 0.

I go deeper and deeper: The product of all reals (0, 1] is 0? The sum of all reals (-1, 1] is 1? The product of all reals (0, 1.0000001] is... unbounded infinite... but definitely > 0? What even is the sum of all reals [-.000001, 1]?

Is there literature, or field of study that I don't know the name of that talks about these things? Or is it as trivial as it sounds at a surface level and nobody cares about the "computability" of them that my mind is hung up on?

Determine if the set A of all intersection points in R2 of the family of lines {y=mx:m∈Z}

with the family of circles{x2+y2=r2:r∈Q} is countable or uncountable.

where should I start, thanks

Why difference is countable in this sentence:

Many nouns can be used as countable or uncountable nouns, usually with a difference in meaning.

Did you hear a noise just now?

I can't work here. There's too much noise.

I guess the first noise mean a particular noise, like a noise made by traffic, etc., the second noise means every noise around you. If that's the case, why not use noises instead?

I think the sentences in the following pairs have the same meanings, but why the ones on the left are countable, the ones on the right aren't:

What a beautiful view! = What beautiful scenery!

It was a good suggestion. = It was good advice.

Why a hair can mean one single hair, but a rice can't mean one single grain of rice?

Why headache is countable?

ex. I have got a headache.

In an effort to create functions that counter the intuition behind what it means to be continuous, I came up with the following question:

Is there a function, f, such that for any interval I⊆D of positive Lebesgue measure (that is, not a closed interval of one number), where D is its domain, the set continuities of f|_I (f restricted to I) is dense in I and the set of discontinuities of f|_I is uncountable and dense in I?

I am specifically looking at single variable functions on the real numbers, though answers about functions on other metric spaces are welcome.

Some examples of intuition-breaking functions that come close to the type of function I am looking for are (for simplicity, take (0,1) as the domain):

Dirichlet’s function (rational indicator function): D(x)=1 if x is rational and 0 otherwise.

This is discontinuous everywhere.

Thomae’s function: T(x) = 1/q if x is rational and x=p/q for some coprime integers p,q. T(x) = 0 if x is irrational

This is continuous only at the irrational numbers (dense, uncountable in every interval) and discontinuous at the rationals (dense, countable in every interval).

Cantor indicator function: C(x) = 1 if x is in the cantor set. C(x) = 0 otherwise.

This is continuous only at numbers not in the cantor set (dense, uncountable in every interval) and discontinuous at the numbers in the cantor set (not dense, uncountable in only some intervals)

Mix of Thomae’s function and the Cantor set indicator function : g(x) = T(x) if x is rational. g(x) = C(x) otherwise.

This is continuous only at numbers that are irrational and not in the cantor set (dense, uncountable in every interval) and discontinuous at numbers that are rational or in the cantor set (dense, uncountable in only some intervals).

Function I found: h(x) = (3^-n -|x-a_n|)D(x) for the smallest n such that |x-a_n| < 3^n, where a_n is some enumeration of the rationals in (0,1). h(x) = 0 if no such n exists.

This is continuous on a set that not dense, and uncountable in only some intervals and discontinuous on a set that is dense and uncountable in all intervals.

I think the requirement of uncountability in every interval can be restated as having positive lebesgue density everywhere (or almost everywhere).

I’ve heard that there is a function with set of discontinuities A iff A is Fσ (a countable union of closed sets), but I haven’t been able to use that to prove or disprove the existence of the type of function I’m looking for.

The answer that I have found elsewhere is that the integers can be listed as such

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13...

and that the same cannot be done for the real numbers between 0 and 1. I don't see why this is true because I can list them as follows.

0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.11, 0.21, 0.31...

Notice that there I have constructed the second list by taking the mirror image of the first list about the decimal point, so there is definitely a 1 to 1 mapping of integers to real numbers between 0 and 1.

The other reason usually given is that if you take a set which is assumed to contain all the real numbers between 0 and 1 then you can create a new number by making the first digit different than the first digit of the first number, the second digit different than the second digit of the second number and so on.

But you can also do this with the integers! Just create a new integer and make different than the first number in the ones place, different than the second number in the 10's place and so on.

Please help me see where I am wrong above.

I think this would be true. Suppose a subset T is uncountable. We can then construct a surjection g from S to T. This means that Card(S) >= Card(T), but T is uncountable. This suggests S is also uncountable.

I recently learned this fact but I am struggling to get an intuition for it. Going off of this definition, it seems to me that this power set would also be countable:

> A set is countably infinite if its elements can be put in one-to-one correspondence with the set of natural numbers. In other words, one can count off all elements in the set in such a way that, even though the counting will take forever, you will get to any particular element in a finite amount of time.

Taking the power set of the natural numbers as an example, I could order its elements this way:

1. Add the unchosen set with cardinality 1 whose element, the natural number x, is smallest.
2. Add the set Y which contains all natural numbers from 1 to x.
3. Add any members of the power set of Y which have not yet been included. This would involve of a finite number of additions to our list, and could be done in a methodical way so as to preserve some computable order.
4. Repeat.

For example, the ordering would begin like this:

{1}, {2}, {1,2}, {3}, {1,2,3}, {1,3}, {2,3}, {4}, {1,2,3,4}, ...

It seems to me that this is an ordering which would include every element of the power set at a specific point in correspondence with the set of natural numbers. What am I getting wrong?

If so, how?

Is there such a thing as an uncountable set that is composed entirely of countable bounded proper subsets, or would that pretty much imply that the set is countable?

I was noodling on scratch paper trying to prove: if you have an ordered set, and every element is a countable "distance" from the left-most or zero element, then the whole set must be countable.

Turns out this is wrong. Consider a well-ordering of the reals, so that's an uncountable set. Then within that ordering, look at the set S of all elements that are an uncountable distance from the zero element. (If S is empty, then the original set is you're counterexample, and you're done.) If S is non-empty, by well-ordering, it must have a least element x. So consider the set T of all elements below x.

By definition of T, every element is a countable distance from the zero element, but since T consists of all the elements less than x and x is in S, T must be uncountable.

Presumably this is a known result, but I couldn't find it with any combination of Googling "uncountable ordered set with every element countable distance from zero" or anything similar. Is there a name for this set?

I know that countable sets have measure 0.

If we reject CH then can we say anything about the measures of sets with cardinality strictly between the naturals and the reals? Are such sets even measurable?

We had this discussion in college. We argued a finite (countable) union of countable sets is also countable, but couldn't came up with a quick proof for infinite unions.

Thanks!

EDIT: it seems that I got a bit lost in translation. Our question was more like: what is "an infinite sum of uncountable unions". Here it's sort of written down: http://imgur.com/gallery/wzQzSzy

So from what I know about the hierarchy of cardinalities, you've got countable, which is equivalent to |N| (or aleph_0, which I don't know how to type), and uncountable, which is equivalent to |R| (or aleph_1, if you accept CH). I've only heard of sets being finite, countable, or uncountable. My question is, does there exist a set with cardinality larger than uncountable? What would be the word for that cardinality?

I understand that |R| = 2^|N|, because the powerset of N can be mapped to the set of infinite binary strings, which can be mapped to R. So would the powerset of R, whose cardinality is 2^|R|, be such a set? or is |R| somehow equal to 2^|R|?

Thanks!

Let me pretext this by saying I'm not a mathematician. According to Cantor, if I understood right, some set of numbers are countable while others are not. The countable set of numbers are those that can be organised (natural and irrational numbers), and the uncountable are those that cannot (real numbers).

My question is, why can't real numbers be counted? Let's take pi, for instance: although the number goes on forever in no logical order, can I not ascribe to every number after the decimal a corresponding natural number, thereby making it structured and countable? I mean, we do this already to refer to specific points of pi (Feynman point at 762^nd decimal place).

Oh, and I was in Halle last month, paying tribute to a true great!

Thanks in advance!

Edit: Seeing as it's customary nowadays on Reddit to edit your submissions, here's mine:
(i) a big thank you to you all;
(ii) it seems Cantor's diagonal argument is what I was after;
(iii) I mistakenly thought Cantor meant any real number within the set of real numbers is uncountable. As SometimesY pointed out, Cantor meant all real numbers.

Thank you again and wish you all a wonderful new year!

For example, what theorems state that (statement X is true) for (something in regards to sets) when these sets have cardinality <= P^k (N) where 1<= k < infinity.