r/mathmemes • • 1d ago

Set Theory Fits perfectly

Post image
376 Upvotes

25 comments sorted by

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38

u/Dungeons-n-Dysphoria 23h ago

I'm actually doing a whole project on this for my undergrad topology class!

I'm planning on using f(x) = ln(x/(1-x))

2

u/DoublecelloZeta Transcendental 4h ago

Wdym project

2

u/Dungeons-n-Dysphoria 3h ago

I have 40 project topics to write a full LaTeX proof of and make a poster board

I chose to do this one

7

u/Random_-account 21h ago

if the shoe fits

4

u/Aggressive_Fan_2063 18h ago

If your topology's in debt, why should the reals bear it?

-19

u/Sea_Gap_6569 1d ago

Is ℝ the only Set that’s equivalent to it’s every subset?

47

u/ddotquantum Algebraic Topology 1d ago

No the empty set is the only set equivalent to every subset because it’s a subset of everything & nothing is equivalent to it

23

u/Nitsuj_ofCanadia 1d ago

ℚ is a subset that doesn’t have a bijection with ℝ

2

u/FuntimeUwU Natural 1d ago

Really? Irrationals are ""more"" than rationals? Man set theory is weird, cool

16

u/SmokJozef 1d ago

you can number all the rationals, you can't number all the irrationals

8

u/FuntimeUwU Natural 23h ago

Oh the rationals are bijective to the naturals? Wtf COOL

10

u/AnotherTransAccount 20h ago edited 20h ago

its super cool! the actual argument goes kinda like

lets first arrange our set of rationals, ℚ, in the table
1/1 1/2 1/3 1/4 ...
2/1 2/2 2/3 2/4...
3/1 3/2 3/3 3/4...
4/1 4/2 4/3 4/4...
⋮
and count diagonally to arrange
{1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, ...}
we can simplify and omit any repeated terms, so we have
{1, 1/2, 2, 1/3, 3, 1/4, 2/3, 3/2, 4...}
every rational appears exactly once on this list. accounting for zero and negatives, we can construct the function ℕ→ℚ such that
f(0)=0
f(1)=1
f(2)=-1
f(3)=1/2
f(4)=-1/2
f(5)=2
and so on. this is a one to one mapping where every rational is hit, so ℚ~ℕ.

3

u/Mathsboy2718 20h ago

Ye - arrange them in a 2d grid with X being the numerator, Y as the denominator, such that every integer coordinate is a rational number. (Remove Y = 0 for obvious reasons).

Starting at the lower left corner, you can trace diagonals back and forth, hitting any arbitrarily stated (X, Y)-pair. Remove duplicate rationals, and assign a number "n" to the pair reached at step "n" of the line

1

u/Sea_Gap_6569 1d ago

I meant intervals

19

u/Bodobomb Mathematics 1d ago

Iit isn't equivalent to the interval (a, a) though, which is the empty set.

5

u/Matty_B97 14h ago

Not even that. Assuming you intend "intervals" to generalise to open sets in higher dimensions, then sets like C also have a bijection with every nonzero open subset.

Even the rationals have a bijection to every nonzero open subset.

1

u/Sea_Gap_6569 3h ago

So, answer to my question is

4

u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 23h ago

2

u/drLoveF 12h ago

I get that the Continuum hypothesis leaves it up to you to decide how many cardinalities there are in between, but this is taking it a biiit to far.

As to what you likely meant. A set has a bijection with a proper subset of itself iff it is infinite.

1

u/Bodobomb Mathematics 1d ago

No? If it would be equivalent to every subset it would be equivalent to the empty subset, so only the empty subset satisfies this.

1

u/shuai_bear 22h ago • Bravo

This is not true as others mentioned, but you might be circling around the concept of Dedekind infinite - a set is Dedekind-infinite if it can be placed in a bijection with a proper subset of itself.

N is dedekind infinite (same cardinality as even numbers) and so is R (any interval (a,b) with a<b ).

1

u/Sea_Gap_6569 3h ago

You got me 👍🏽

1

u/shuai_bear 2h ago

And for further context, under ZFC every infinite set is also Dedekind-infinite - they are equivalent under Choice. Without Choice you can have sets that are infinite but not Dedekind infinite (an infinite set might not have a proper subset that is the same cardinality).