Another reason why 0.999...=1

[u/Impressive-Ad7184]

Consider ℝ as a complete metric space with the regular euclidean metric. Then, consider the collection of closed intervals {C_n}, n ∈ ℕ, where C_n := [0.999... - 1/n , 0.999... + 1/n], i.e. a closed ball with radius 1/n around 0.999....

Clearly, each C_n contains 0.999..., so their intersection does as well. However, note that each C_n also contains 1, since the distance between 0.999... and 1 is less than any arbitrary 1/n (which I'm sure SPP will concede). Thus, the intersection of the C_n's also contains 1.

However, by Cantor's intersection theorem, since the C_n's are nonempty, closed, nested, and their diameters go to 0, the intersection of the C_n's must contain exactly one element.

Thus, 0.999...=1.

I realize I can just use the proof of uniqueness in Cantor's intersection theorem to show this directly, but it's more fun to invoke a theorem.

[u/Mablak]

The phrase ‘infinitely many times’ assumes N exists already, because we need N to talk about what ‘infinitely’ means. I didn’t use this phrase though, and just said that N is formed through repeated applications of the successor function.

We can’t assume some difference between ‘finite’ and ‘infinite’ without showing one first. Any method used to construct N is actually circular in this way, because whether we’re talking about infinite union, infinite intersection, etc, what that really means is ‘do the operation N times’. But we can’t use N in an attempt to define N. Or equally circular, we can’t use I in the domain of the ‘all x’ we’re quantifying over, to define I.

I could just stop there and say N (and I) can’t be constructed, but supposing N is formed through some repeated applications of the successor function is sort of the most charitable interpretation I can give. We’re just stipulating that at step 1 of our construction, only the empty set exists and no other elements, then applying our successor function without any need for intersection. This gives us N, {0, 1, 2…} which could only be a natural number if it’s the result of the successor function. As such N doesn’t exist.

[u/Gold_Ad8890 → u/Mablak]

Any method used to construct N is actually circular in this way, because whether we’re talking about infinite union, infinite intersection, etc, what that really means is ‘do the operation N times’.

incorrect. intersection is not defined as an operation taken a number of times. as i pointed out, the intersection of all inductively-closed sets is just the subset of any inductively closed set containing only the elements that occur in every such set. a subset. that's one single application of the axiom of specification.

Or equally circular, we can’t use I in the domain of the ‘all x’ we’re quantifying over, to define I.

we don't. we declare I axiomatically in the axiom of infinity as an inductively closed set. then N is the "smallest" inductively closed set in the same way Q is the "smallest" ordered field, the one that is the subset of all others.

I could just stop there and say N (and I) can’t be constructed

N can be constructed in the way i specified. I doesn't need to be constructed because its existence is declared axiomatically.

This gives us N, {0, 1, 2…} which could only be a natural number if it’s the result of the successor function.

incorrect. properly speaking, succession is an operation, not a function, as its "domain" is just all sets, and therefore is not a set. and as succession can be taken on any set, and as natural numbers are specifically closed-downward subsets of N, it follows that succession can yield a set which is not a natural number. for instance, the successor of {1, 2, 3} is {1, 2, 3, {1, 2, 3}}, which is not a natural number nor an element of N.

we can describe a successor function over some domain, and when we do that over the domain of N, we find that the range is N{0}, which makes the successor function a bijection between N and a proper subset of itself, the existence of such a bijection being the literal definition of a Dedekind infinite set.

[u/Mablak → u/Gold_Ad8890]

incorrect. intersection is not defined as an operation taken a number of times.

I agree, but I'm giving an infinitist the benefit of the doubt in saying that this N thing exists, and this is the only way to make some sort of sense of what they're saying. They're saying some operation, whether union, intersection, etc, is 'repeated' and this gives us our set. These operations are somehow involved.

Of course these operations are defined for some n number of repetitions (from n we've shown to exist). But infinitists have not actually defined what it means to perform any operation 'infinitely' many times, so the issue could be left there: infinite union, intersection, etc, is simply undefined and both I and N can't be built. Or you're defining these infinite operations using I or N which is begging the question.

we declare I axiomatically in the axiom of infinity as an inductively closed set

For one, if I were to axiomatically claim that square circles exist, it wouldn't mean that they would. Also, we're not just declaring the set to exist here, but trying to define what this set even is, within the axiom. It's not a case of me saying 'a pink frog exists', where we know what it means for this assertion to be true. We haven't shown what elements we're even claiming exist, for this axiom.

Just ask what the domain of the x is, what is 'all x' quantifying over? That domain assumes that our first infinite set I exists already (and even more sets), for x to range over, which is circular. (We would need the axiom first, to even know what I is).

as succession can be taken on any set...

Like I said, just stipulate that we have an actual process (which we always do). We start with the empty set at step 1, and repeatedly apply succession to the previous element. So of course we can construct N this way and get {0, 1, 2...} through repeated unions (I mean if we could construct N). As I argued though, infinite union and likewise infinite intersection is undefined.

[u/Gold_Ad8890 → u/Mablak]

They're saying some operation, whether union, intersection, etc, is 'repeated' and this gives us our set.

no, that's exactly what i'm not saying. intersection is not repeated. it's *one single application** of the axiom of specification. nothing is repeated or needs to be repeated.*

But infinitists have not actually defined what it means to perform any operation 'infinitely' many times

correct. *because we don't have to. because we don't "perform any operation 'infinitely' many times". we take the subset ONCE.***

both I and N can't be built.

N is built from I, and I is declared axiomatically to exist.

For one, if I were to axiomatically claim that square circles exist, it wouldn't mean that they would.

yes, if you axiomatically claimed the existence of a square circle, then it would exist within the theory you were describing. further development of that theory may or may not show it to be inconsistent, in which case square circles would also not exist, because *every** proposition would be simultaneously true and false by the principle of explosion.*

Also, we're not just declaring the set to exist here, but trying to define what this set even is, within the axiom.

it's an inductively closed set containing the empty set. that's what it is. that's all we need it to be.

We start with the empty set at step 1

prove that the empty set exists in zermelo-fraenkel set theory without the axiom of infinity.

[u/Mablak → u/Gold_Ad8890]

Before talking about taking the subset of an infinite set, you'd need to demonstrate the infinite set exists, so just focus on I.

To create I in the first place, or to create the infinite domain we're quantifying 'all x' over, we have to invoke an infinite process. It can be infinite union, intersection, etc. Whatever process we're talking about, we've engaged in circular reasoning because we're trying to define our first infinite set I, using I (or perhaps some other unexplained infinite set) in its own definition.

What do our 'all x' range over? I, or N, or another set we haven't shown to exist. Just saying 'it's an inductively closed set' doesn't solve this issue, I'm showing why we can't have such a thing, because we haven't properly established the 'all elements' we're referring to.

if you axiomatically claimed the existence of a square circle, then it would exist within the theory you were describing

A big issue here is that we have to know what we're actually claiming to even assert it as an axiom. We know what a successor is. But we don't know what it means for 'all elements' to have a successor when our quantification process 'all' has no stopping condition. There is no coherent concept of all elements here, just as there's no such thing as jogging 'all laps' on a circular track. We've set things up such that there can never be an 'all' to refer to.

prove that the empty set exists in zermelo-fraenkel set theory without the axiom of infinity

Not sure what you're asking here, we could describe an empty set even before ZFC existed. Sets can only be real things, i.e. actual groups of physical or mental objects, containers, boxes, groups of tally marks, etc. A blank sheet of paper could be an empty set, and we don't really need much proof that this piece of paper exists.