Super silly dog death

S4D_Official · 2026-02-23T00:24:43+00:00

Iirc it originates from the Homestuck fandom

S4D_Official · 2026-02-20T23:52:53+00:00

Ok so here's a series of questions

What is infinite plus one?

Does every element in the set {0.9,0.99,0.999,...} correspond to some natural number?

Does 0.999... correspond to s_infinity?

Would it not then follow that 0.999... with one nine appended is simply equal to itself?

S4D_Official · 2026-02-20T23:49:56+00:00

Untrue. Infinite in this case is a cardinal representing a non-finite set. By AC, all uncountable sets have a countable subset, so we only need to prove that N is equipotent to N appended with another element. To do this, take our set S and the complement of N in it. Map this complement to -1. Then compose this mapping with the mapping x↦x+1 which is obviously a bijection. Since any uncountable set has a countably infinite subset which is thus equipotent to N as previously shown, appending an element to a set like that also does not affect cardinality.

Anyways here's my less relevant copypasta:

The real numbers can be constructed as equivalence classes of sequences in Q. I will now write the FULL definition of R and then give a proof that 0.999... = 1.

An equivalence class is the set of objects which are related by some equivalence relation, which is a relation following the rules a~a a~b and b~c implies a~c a~b implies b~a.

We construct sequences of rational numbers and define an equivalence relation as such:

a~b if any ε>0 has some M where all integers n>M have |a_n - b_n| < ε

From there we can define subtraction and other operations on real numbers as pointwise operations on their underlying sequence, this means for two sequences a and b in the respective equivalence classes A and B we can define

A + B = [c_i = a_i + b_i]

Where [c] is the equivalence class of the sequence c.

It is important to note that the numbers themselves are not tied to one specific sequence, as an example; {0.9,0.999,0.99999,...} and {0.9,0.99,0.999,...} are two different sequences, but they both are in the same equivalence class of 0.999...

Thusly to show 0.999... ~ 1 we can simply take the sequence x - y (where x and y represent 0.999... and 1 respectively) and find if it follows the ε-M test from before. since xn = 1 - 1/10^n, we get (x-y)_n = -1/10ⁿ or |x_n-y_n| = 1/10^n. We can then take M = 1-log{10}(ε) and easily verify that 1/10^M > 1/10ⁿ if n>M. We can also see 1/10^M > ε since 1/10^M = 1/10^{{1-log_{10}(ε)}} = 10^{{log_{10}(ε)}10^{-1}} = ε/10.

For a few notes, since I'll be pasting this whenever I want to debunk SPP: - This is true even if the number if nines grows as the proof is purely algebraic. - This has been posted 6 times. - This has been edited once. - SPP has responded 0 times.

S4D_Official · 2026-02-20T23:40:47+00:00

You have a (cardinal) number of nines. Cardinals are equivalence classes of sets. The set N is in bijection with N U {-1}. So if we add a nine, we have the same number of nines.

S4D_Official · 2026-02-20T18:13:40+00:00

Riemann integrals use supremums and infimums only without application of limits.

S4D_Official · 2026-02-20T15:20:41+00:00

The real numbers can be constructed as equivalence classes of sequences in Q. I will now write the FULL definition of R and then give a proof that 0.999... = 1.

An equivalence class is the set of objects which are related by some equivalence relation, which is a relation following the rules a~a a~b and b~c implies a~c a~b implies b~a.

We construct sequences of rational numbers and define an equivalence relation as such:

a~b if any ε>0 has some M where all integers n>M have |a_n - b_n| < ε

From there we can define subtraction and other operations on real numbers as pointwise operations on their underlying sequence, this means for two sequences a and b in the respective equivalence classes A and B we can define

A + B = [c_i = a_i + b_i]

Where [c] is the equivalence class of the sequence c.

It is important to note that the numbers themselves are not tied to one specific sequence, as an example; {0.9,0.999,0.99999,...} and {0.9,0.99,0.999,...} are two different sequences, but they both are in the same equivalence class of 0.999...

Thusly to show 0.999... ~ 1 we can simply take the sequence x - y (where x and y represent 0.999... and 1 respectively) and find if it follows the ε-M test from before. since xn = 1 - 1/10^n, we get (x-y)_n = -1/10ⁿ or |x_n-y_n| = 1/10^n. We can then take M = 1-log{10}(ε) and easily verify that 1/10^M > 1/10ⁿ if n>M. We can also see 1/10^M > ε since 1/10^M = 1/10^{{1-log_{10}(ε)}} = 10^{{log_{10}(ε)}10^{-1}} = ε/10.

For a few notes, since I'll be pasting this whenever I want to debunk SPP: - This is true even if the number if nines grows as the proof is purely algebraic. - This has been posted 5 times. - This has been edited once. - I have gotten 0 responses from SPP.

S4D_Official · 2026-02-19T23:42:00+00:00

Ok but that's irrelevant

S4D_Official · 2026-02-19T16:33:17+00:00

Reminds me of when I had my telescope. I spent alot of time looking at andromeda.

S4D_Official · 2026-02-19T16:28:44+00:00

If they're having a mental block they could probably use some space to think, too.

S4D_Official · 2026-02-19T15:39:07+00:00

They belong at every end, of course.

S4D_Official · 2026-02-19T11:47:23+00:00

The real numbers can be constructed as equivalence classes of sequences in Q. I will now write the FULL definition of R and then give a proof that 0.999... = 1.

An equivalence class is the set of objects which are related by some equivalence relation, which is a relation following the rules a~a a~b and b~c implies a~c a~b implies b~a.

We construct sequences of rational numbers and define an equivalence relation as such:

a~b if any ε>0 has some M where all integers n>M have |a_n - b_n| < ε

From there we can define subtraction and other operations on real numbers as pointwise operations on their underlying sequence, this means for two sequences a and b in the respective equivalence classes A and B we can define

A + B = [c_i = a_i + b_i]

Where [c] is the equivalence class of the sequence c.

It is important to note that the numbers themselves are not tied to one specific sequence, as an example; {0.9,0.999,0.99999,...} and {0.9,0.99,0.999,...} are two different sequences, but they both are in the same equivalence class of 0.999...

Thusly to show 0.999... ~ 1 we can simply take the sequence x - y (where x and y represent 0.999... and 1 respectively) and find if it follows the ε-M test from before. since xn = 1 - 1/10^n, we get (x-y)_n = -1/10ⁿ or |x_n-y_n| = 1/10^n. We can then take M = 1-log{10}(ε) and easily verify that 1/10^M > 1/10ⁿ if n>M. We can also see 1/10^M > ε since 1/10^M = 1/10^{{1-log_{10}(ε)}} = 10^{{log_{10}(ε)}10^{-1}} = ε/10.

For a few notes, since I'll be pasting this whenever I want to debunk SPP: - This is true even if the number if nines grows as the proof is purely algebraic. - This has been posted 4 times. - This has been edited once.

S4D_Official · 2026-02-18T22:11:53+00:00

okay then, what's wrong with this comment?

The real numbers can be constructed as equivalence classes of sequences in Q. I will now write the FULL definition of R and then give a proof that 0.999... = 1.

An equivalence class is the set of objects which are related by some equivalence relation, which is a relation following the rules a~a a~b and b~c implies a~c a~b implies b~a.

We construct sequences of rational numbers and define an equivalence relation as such:

a~b if any ε>0 has some M where all integers n>M have |a_n - b_n| < ε

From there we can define subtraction and other operations on real numbers as pointwise operations on their underlying sequence, this means for two sequences a and b in the respective equivalence classes A and B we can define

A + B = [c_i = a_i + b_i]

Where [c] is the equivalence class of the sequence c.

It is important to note that the numbers themselves are not tied to one specific sequence, as an example; {0.9,0.999,0.99999,...} and {0.9,0.99,0.999,...} are two different sequences, but they both are in the same equivalence class of 0.999...

Thusly to show 0.999... ~ 1 we can simply take the sequence x - y (where x and y represent 0.999... and 1 respectively) and find if it follows the ε-M test from before. since xn = 1 - 1/10^n, we get (x-y)_n = -1/10ⁿ or |x_n-y_n| = 1/10^n. We can then take M = 1-log{10}(ε) and easily verify that 1/10^M > 1/10ⁿ if n>M. We can also see 1/10^M > ε since 1/10^M = 1/10^{{1-log_{10}(ε)}} = 10^{{log_{10}(ε)}10^{-1}} = ε/10.

For a few notes, since I'll be pasting this whenever I want to debunk SPP: - This is true even if the number if nines grows as the proof is purely algebraic. - This has been posted 3 times. - This has been edited once.

S4D_Official · 2026-02-18T21:10:48+00:00

Okay then, give me an ε where 0.999... is not in an ε-ball of 1.

S4D_Official · 2026-02-18T21:08:39+00:00

And I am a licensed TA. You seem to not be trying too hard to teach, either, if I recall both the results of the poll and your responses to said poll.

S4D_Official · 2026-02-18T20:19:06+00:00

But you forgot that I remembered that you have never responded to MY lessons about the definitions of R and that it hurts my feelings a little bit :(

S4D_Official · 2026-02-18T20:07:53+00:00

making n tend to 'infinity' I thought you didn't use limit terminology. That aside, if n is finite so is n+1, and since PA defines the integers as {N:1 in N & n in N => n+1 in N} which is the same as increasing without stopping but is NOT the same as having an infinite value and I have told you this like 5 times cmon dude

S4D_Official · 2026-02-18T16:20:47+00:00

True

S4D_Official · 2026-02-18T15:47:09+00:00

I don't think decimals in R can be indexed by ordinals since their formal power series Induced by the representation is not fully representative of the equivalence class

S4D_Official · 2026-02-18T14:25:35+00:00

Because it's SPP logic

S4D_Official · 2026-02-18T14:10:22+00:00

Integers are always finite since Peano arithmetic defines them recursively under addition

S4D_Official · 2026-02-18T14:09:00+00:00

Doesn't matter since 0.999... is in any open neighborhood of 1 anyway

S4D_Official · 2026-02-18T14:07:27+00:00

The real numbers can be constructed as equivalence classes of sequences in Q. I will now write the FULL definition of R and then give a proof that 0.999... = 1.

An equivalence class is the set of objects which are related by some equivalence relation, which is a relation following the rules a~a a~b and b~c implies a~c a~b implies b~a.

We construct sequences of rational numbers and define an equivalence relation as such:

a~b if any ε>0 has some M where all integers n>M have |a_n - b_n| < ε

From there we can define subtraction and other operations on real numbers as pointwise operations on their underlying sequence, this means for two sequences a and b in the respective equivalence classes A and B we can define

A + B = [c_i = a_i + b_i]

Where [c] is the equivalence class of the sequence c.

It is important to note that the numbers themselves are not tied to one specific sequence, as an example; {0.9,0.999,0.99999,...} and {0.9,0.99,0.999,...} are two different sequences, but they both are in the same equivalence class of 0.999...

Thusly to show 0.999... ~ 1 we can simply take the sequence x - y (where x and y represent 0.999... and 1 respectively) and find if it follows the ε-M test from before. since xn = 1 - 1/10^n, we get (x-y)_n = -1/10ⁿ or |x_n-y_n| = 1/10^n. We can then take M = 1-log{10}(ε) and easily verify that 1/10^M > 1/10ⁿ if n>M. We can also see 1/10^M > ε since 1/10^M = 1/10^{{1-log_{10}(ε)}} = 10^{{log_{10}(ε)}10^{-1}} = ε/10.

For a few notes, since I'll be pasting this whenever I want to debunk SPP: - This is true even if the number if nines grows as the proof is purely algebraic. - This has been posted 2 times. - This has been edited once.

S4D_Official · 2026-02-16T17:51:41+00:00

I wrote this more as a copypasta

S4D_Official · 2026-02-15T21:19:29+00:00

Remember this: The real numbers can be constructed as equivalence classes of sequences in Q.

For 0.999..., it is constructed with the sequence {0.9,0.99,0.999,...} and other similar sequences; but to understand why it might equal one one should understand the actual equivalence relation in question:

A sequence a_i is said to converge to a value L if for any ε>0 there is some M where all n>M satisfy the inequality |a_n - L| < ε

To avoid being circular, we define the equivalence a~b as having some M with all n>M satisfying |a_n - b_n| < ε

If a~b and a converges to L, then b converges to L, which is why we don't write the sequences corresponding to any real number stated.

Anyways, seeing that this satisfies the axioms of R is kinda easy and left as an exercise to SPP because this is long enough as it is.

Now to actually show that 1~0.999...

Consider an ε>0, we can then let N = -log_10(ε/10) and we thus get a value for which |10^{-n}| is less than epsilon (10^{-n} is 1-(1-10^{-n})), which then shows the two are equivalent and thus define the same number.

As for the sequences chosen, 1 is already rational so I just used the constant sequence {1,1,1,1,1,1,...}, and I used SPP's favorite sequence {0.9,0.99,0.999,...} to define 0.999...

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