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The general enterprise of calibrating the strength of classical mathematical theorems in terms of the axioms, typically of set existence, needed to prove them; originated in its modern form in the 1970s by H. Friedman and S. G. Simpson (see R.A. Shore, "Reverse Mathematics: The Playground of Logic", 2010).
2
votes
Accepted
Proving boundedness of continuous images of [0,1] in WKL0
Here is a sketch of how the proof might go.
If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S_n\ne\emptyset$ for all …
25
votes
Accepted
What is the reverse mathematical strength of the fundamental theorem of algebra?
Tanaka and Yamazaki (in the volume Reverse Mathematics 2001, see review) show that a substantial portion of field theory can be done in the weak base theory RCA$_0$, by proving in RCA$_0$ the fundamen …
20
votes
Accepted
Why is weak Kőnig's lemma weaker than Kőnig's lemma?
The issue is that for a finitely branching subtree $T$ of $\omega^{<\omega}$, the function $f$ mapping $\sigma$ to the greatest $n$ such that the concatenation $\sigma ^\frown n$ is in $T$ may not be …
10
votes
Proof complexity of two directions of equivalency?
This fits in the program of reverse mathematics. For instance: a subtree of the set of all finite binary strings has an infinite path iff it is infinite. One direction is provable in RCA $_0$ and the …
6
votes
Accepted
Is 0' of PA degree relative to a non-low set?
No, by the Arslanov completeness criterion $0'$ is only DNC (Diagonally non-computable) relative to low sets. And PA implies DNC.
8
votes
Accepted
Reverse Math of High Sets?
In the paper "On a conjecture of Dobrinen and Simpson regarding almost everywhere domination", Binns, Lerman, Solomon and I constructed $\omega$-models of this "high" principle which demonstrate it do …
18
votes
Why is this new result such a big deal?
They show that $\DeclareMathOperator{\WKL}{WKL}\DeclareMathOperator{\RT}{RT}\DeclareMathOperator{\RCA}{RCA} \RT^2_2$ is $\Pi^0_3$-conservative over $\RCA_0$. Thus, there is no way that $\RT^2_2$ can b …