Is the set difference of two different limit ordinals always infinite Assume $β_1$ and $β_2$ are limit ordinals and $β_1 < β_2$. Is it true that ($β_2$ \ $β_1$) is infinite.
My intuition says that it has to be infinite, because if it's finite and |$β_2$ \ $β_1$| $= n$, then $β_1 + n = β_2$ and therefore $β_2$ is not limit. I cannot prove it formally and I'm not sure if my reasoning is correct.
 A: Your intuition is correct.
There is an even stronger statement, the set difference between a limit ordinal, and any if its elements is infinite.
Here I will show your claim, but it can be easily generalized.
Consider the set:
A = {n∈ω|($β_1$+n)∈$β_2$}
Claim 1 (verify yourself by induction): A contains every positive natural number.
Claim 2 (verify yourself ): For all n>0, there exists a unique $α_n$ s.t. $α_n$ = ($β_1$+n)∈$β_2$.
Hint: Existence is given by Claim 1, for uniqueness assume $α_m$ = $α_n$ show that m = n
By Replacement Axiom:
B = {$α_n$|n∈ω and n > 0}  exists.
Notice that B is infinite, as it has the same cardinality as ω. You should verify this for yourself if you are unsure about the claim.
Claim 3 (you know how this goes) Verify:
B ⊆ $β_2$/$β_1$
Since B is infinite, and it is a subset of $β_2$/$β_1$, then $β_2$/$β_1$  is infinite.
It is a common assigment to show that the superset of an infinite set is infinite. If you haven't seen such a proof before, I reccomend you verify it yourself.
