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What mathematics can be built with standard ZFC with Axiom of Infinity replaced with its negation?

Can the analysis be built? Is there special name for "ZFC without Infinity" set theory?

I also assume that using symbols like $+\infty$, $-\infty$ when dealing with properties of functions and their limits would still be possible even with Axim of Infinity negated (correct me if I am wrong).

UPDATE

In light of the answer by Andres Caicedo which suggested Peano arithmetic, I want to point out that Wikipedia says about Peano arithmetic "Axioms 1, 6, 7 and 8 imply that the set of natural numbers is infinite, because it contains at least the infinite subset". I do not know how this can be interpreted as having axiom of infinity but I am interested what if Peano arithmetic modified the following way:

  • Added an axiom 5a:

There is a natural number $\infty$ which has no successor, for any natural number n $S(\infty)=n$ is false.

  • Axiom 6 modified:

For every natural number n except $\infty$, S(n) is a natural number

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    Not completely relevant, but related nonetheless: http://math.stackexchange.com/questions/77992/surreal-numbers-without-the-axiom-of-infinity2012-02-09
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    "ZFC with infinity" would mean ZFC with the infinity axiom removed. If you attach the negation of the infinity axiom, you are assuming that there are no inductive sets (I wonder if that is enough to prove all sets are finite?) Also, I imagine that the axiom of choice in this context is redundant.2012-02-09
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    @Bill, if we assume that there is no inductive set but have Choice, we can still prove that every set is equinumerous to an ordinal. Since $\omega$ is assumed not to exist, this means that every set must be finite. I think mathematical induction arguments could still be carried out, in the guise of "transfinite" induction, so I would expect one could prove that every set is Dedekind-finite too.2012-02-09
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    @HenningMakholm nice! +12012-02-10
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    For the proposed "5a" axiom, it is not compatible with the usual language of PA. $S$ is a function symbol, so $S(x)$ will be a term in need of an interpretation for every element of your model, including $\infty$. You can change to a language using a two-place predicate $S(x,y)$ that one thinks of as meaning "$y$ succeeds $x$." But it's not something you can straightforwardly add.2016-07-04

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