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May 17

Right Inverse Needs Choice?

I'm looking at a practice problem that seems to demand more than it should. Suppose f is a function (for concreteness say from R to R), then a function g is called a right inverse for f if f(g(y))=y for all y. Prove that if f is surjective then it has a right inverse. The preimages of f form a partition, and the right inverse forms a choice function, so this seems like it requires the axiom of choice to be true. Am I missing something? — Preceding unsigned comment added by 162.196.140.139 (talk) 16:42, 17 May 2014 (UTC)[reply]

You're quite right, saying that every surjective function has a right inverse is equivalent to the axiom of choice. Dmcq (talk) 17:28, 17 May 2014 (UTC)[reply]

Also see Axiom of choice #Equivalents. --200.1.109.162 (talk) 23:05, 17 May 2014 (UTC)[reply]

An article with that in and listed as an equivalent! - I suppose that's pretty definitive. Anyway I think the OP should be give extra credit for their response to the assignment. :) Dmcq (talk) 23:50, 17 May 2014 (UTC)[reply]

I think I might need to make a tiny adjustment: the axiom of choice is only needed where you're dealing infinite sets. Otherwise I think you're good without it. RomanSpa (talk) 15:15, 18 May 2014 (UTC)[reply]

What you really need is for the domain to be well-ordered (so you can pick the least pre-image); the axiom of choice implies that all sets are. --Tardis (talk) 13:50, 21 May 2014 (UTC)[reply]

I don't know the context of the assignment, but ZFC is pretty popular, so if it's in a naive set theory context, it seems reasonable to just assume AC is true, just like any ZF axiom is assumed true. -- Meni Rosenfeld (talk) 12:39, 19 May 2014 (UTC)[reply]

New kind of infinity

Not sure where to place this.

http://dubai-computer-services.com/articles/infinities_by_khawar_nehal_19_mar_2014-1.pdf

If you can guide me as to where this can go, then I can place it there.

Regards,

Khawar Nehal — Preceding unsigned comment added by 94.203.215.252 (talk) 17:39, 17 May 2014 (UTC)[reply]

Where to place it? Nowhere. Wikipedia articles are based on published reliable sources - it is not a platform for the promotion of original research. AndyTheGrump (talk) 17:45, 17 May 2014 (UTC)[reply]

Nowhere? That's unduly harsh. Why not on Khawar Nehal's personal website? —Tamfang (talk) 08:21, 18 May 2014 (UTC)[reply]

You may like to study our article ordinal number about different infinities. Outside wikipedia you may use www.academia.edu. Good luck! Bo Jacoby (talk) 17:54, 17 May 2014 (UTC).[reply]