From owner-qed Tue Nov 1 03:03:58 1994 Received: from localhost (listserv@localhost) by antares.mcs.anl.gov (8.6.4/8.6.4) id DAA26502 for qed-out; Tue, 1 Nov 1994 03:02:12 -0600 Received: from lapsene.mii.lu.lv (root@lapsene.mii.lu.lv [159.148.60.2]) by antares.mcs.anl.gov (8.6.4/8.6.4) with SMTP id DAA26497 for ; Tue, 1 Nov 1994 03:02:02 -0600 Received: from sisenis.mii.lu.lv by lapsene.mii.lu.lv with SMTP id AA24175 (5.67a8/IDA-1.4.4 for ); Tue, 1 Nov 1994 11:01:54 +0200 Received: by sisenis.mii.lu.lv id AA28879 (5.67a8/IDA-1.4.4 for QED discussions ); Tue, 1 Nov 1994 11:01:52 +0200 Date: Tue, 1 Nov 1994 11:01:50 +0200 (EET) From: Karlis Podnieks To: QED discussions Subject: Semantics Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Sender: owner-qed@mcs.anl.gov Precedence: bulk Dear Colleagues: (I'm soory for my poor English.) Freiling's axiom is wonderful, but is it the only reasonable extension of Zermelo-Fraenkel's axioms? Is it powerful enough to solve many problems known as undecidable in ZF or ZFC? Two powerful extensions of ZF are well known: - the axiom of constructibility (known as V=L, i.e. all sets are constructible), - the axiom of determinateness (AD). Most writers reject V=L as a proper foundation for set theory because of its "non evidence". I think, this is a somewhat strange position. For me, V=L is some form of the Church's thesis. Have You ever seen a non-constructible set? If not, let us assume that V=L. Why not? Is the axiom of choice "evident"? Let it be evident, if some people like so. But You can demonstrate that AD is "evident" also: For any set A of sequences of natural numbers: Ex[0]Ax[1]Ex[2]Ax[3]Ex[4]... (sequence x belongs to A) or Ax[0]Ex[1]Ax[2]Ex[3]Ax[4]... (sequence x does not belong to A) Now convert the second member of disjunction as follows: not(Ex[0]Ax[1]Ex[2]Ax[3]Ex[4]... (sequence x belongs to A). I.e. AD is "equivalent" to the law of the excluded middle, and therefore, "evident". A wonderful situation for a determined syntacticist (like me): the axiom of choice is "evident", AD is "evident" also, but these two postulates contradict each other. Hence, You can extend ZF axioms in many ways obtaining different wonderful new set theories (call your own extension "evident", if You like so). I agree that "it is too narrow a view of mathematics to consider it as only concerning the consequences of an initially chosen set of axioms". This is only the first (and the main) component of mathematics. The second one is the art of inventing interesting, powerful, wonderful etc. sets of axioms. Karl Podnieks, Dr.Math., determined syntacticist, podnieks@mii.lu.lv Institute of Mathematics and Computer Science University of Latvia Raina Blvd. 29 Riga, LV-1459 Latvia P/S. I would like to be in the member list of QED. Please, let me know, how can I achieve this.