I looked on the web to search for weakly compact cardinals. The web sources indicate many properties of weakly compact cardinals and say that their existence is not entailed by the axioms of ZFC. However it is not indicated whether their existence is {\it consistent} with the axioms of ZFC.
So my question is: Is it known that the axiom of existence of weakly compact cardinals consistent with ZFC?
Thanks
As weakly compact cardinals are in particular (strongly) inaccessible, it follows from Gödel's Second Incompleteness Theorem that ZFC cannot prove the implication "Con(ZFC) implies Con (ZFC + $\exists$weakly-compact )." (Unless, of course, if ZFC is itself inconsistent, at which point this is all a lot of bunk).
(The linked Wikipedia article outlines the basic reasoning.)
Any argument for the consistency of "ZFC + $\exists$weakly-inaccessible" must therefore transcend ZFC. (But, as mentioned by Asaf, no contradictions have been found thus far been under the assumption of the existence of weakly-compact cardinals.)
The weakly compact cardinals are fairly low in the large cardinal hierarchy, just a few skips beyond the inaccessible and Mahlo cardinals, and so one can prove the existence of weakly compact cardinals from any of the stronger large cardinal hypotheses (see Cantor's Attic). For example, the weakly compact cardinals have a strength well below the indescribable cardinals, the unfoldable cardinals, the ethereal cardinals, the subtle cardinals, the ineffable cardinals, and these are all significantly below the Ramsey cardinals and the measurable cardinals, traditionally considered a gateway to the upper class of large cardinals above.
All of these stronger large cardinal notions imply the outright existence of weakly compact cardinals, as well as the consistency of the existence of many weakly compact cardinals. Indeed, this phenomenon is a dominant feature of the large cardinal hierarchy, where the existence of a higher cardinal generally implies the existence of many instances of the lower cardinals. For example, every measurable cardinal $\kappa$ is the $\kappa^{th}$ weakly compact cardinal; every weakly compact cardinal $\gamma$ is the $\gamma^{th}$ Mahlo cardinal; every Mahlo cardinal $\delta$ is the $\delta^{th}$ inaccessible cardinal. In particular, the existence of a higher large cardinal implies the consistency of ZFC with the existence of many of the lower cardinals. If there is a weakly compact cardinal $\gamma$, for example, then the universe $V_\gamma$ up to $\gamma$ satisfies ZFC plus the assertion that there is a proper class of Mahlo cardinals.
It follows (as in Arthur's answer) that we cannot prove the existence of a large cardinal in ZFC or even in ZFC plus a lower large cardinal notion (unless that theory is inconsistent), since this would violate the incompleteness theorem for this lower theory. So even ZFC plus a proper class of hyper-Mahlo cardinals, if consistent, does not suffice to prove the existence of a weakly compact cardinal, precisely because if $\kappa$ is weakly compact, then $V_\kappa$ is a model of ZFC plus a proper class of hyper Mahlo cardinals.
Some might object that this would seem to make the study of large cardinals a doubtful activity, for not only have we failed to prove that the large cardinals exist, we haven't even proved that their existence is consistent, and indeed, we have even proved that we cannot consistently prove that their existence is consistent! Shouldn't this put us off the subject of large cardinals?
No. The point is that because of the incompleteness theorem, we know that there is a hierarchy of consistency strength, a tower of theories each of which implies the consistency of weaker theories in the tower. We wanted to find such a tower of theories, with the property that the consistency of the weaker theories does not prove the consistency of the stronger theories. How fortunate that the large cardinal hierarchy exhibits exactly the features we sought! Furthermore, the large cardinal hierarchy exhibits this tower of consistency strength not by means of weird self-referential convoluted logic statements, as in the incompleteness theorem, but rather with highly natural statements involving infinite combinatorics, such as the existence of measures and considerations of graph colorings. These were questions in which we were already independently interested. Subsequent study of the large cardinal hierarchy has revealed it to be a unifying explanatory force in the nature of set-theoretic truth. (But still, we must be alert to the possibility of inconsistency.)
So it is not a flaw but rather a feature that we cannot prove the consistency of the existence of any of these large cardinals, except from even stronger ones.
Meanwhile, let me point out in answer to Qfwfq's comment on the question, that the consistency of ZFC easily proves the consistency of ZFC + there is no weakly compact cardinal. To see this, let $M$ be any model of ZFC. If it has no weakly compact cardinal, then we're done. If it does, let $\kappa$ be the least weakly compact cardinal of $M$, and observe that the cut-off universe $V_\kappa^M$ satisfies ZFC and has no weakly compact cardinals. Thus, it is relatively consistent with ZFC that there are no weakly compact cardinals.
Consider the axiom of infinity in the axioms of ZF. You cannot prove it holds from the other axioms of ZF; and assuming it means that you can prove the consistency of ZF-Infinity by taking the hereditarily finite sets as a model. Note that ZF-Infinity can formulate enough about the natural numbers for this to be a non-trivial issue.
Large cardinals are often called strong infinity axioms. They are like the axiom of infinity, without such axioms you have a theory of a certain strength, but adding axioms like "There is a weakly-compact cardinal" or "There is a class of Woodin cardinals" gives us a stronger theory, from which we can prove the consistency of weaker theory (e.g. ZFC).
Equally, I could ask you if you can prove that ZFC itself is consistent. The answer depends on your meta-theory. If you assume ZFC+"There is a weakly compact cardinal" then the answer is yes. If you work within the confines of ZFC then the answer is no.
So if we assume the existence of two weakly compact cardinals, we can prove the consistency of ZFC+"There is a weakly compact cardinal", but if we only assume that there is one weakly compact cardinal, it is impossible to prove that for the this theory is consistent -- just like we cannot prove the consistency of ZFC from itself.
If it's any consolation, I am not aware of any contradiction found with weakly compact cardinals yet.
