So I was thinking about the statement, "This statement is false". Now I have heard people say that the truth of this is unprovable. But I would say that it is just nonsensical! I.e that it creates in itself a small paradoxical world.

So maybe I am unfamiliar with what unprovable means!

I.e in ZF set theory what is meant by "It is unprovable to show whether or not there exists a set that contains $\mathbb{N}$, and is contained in $\mathbb{R}$ but is not isomorphic to either $\mathbb{N}$ or $\mathbb{R}$?

This is a fascinating idea for me, and I would very much like to understand it. So if you could also include in you answer some good texts on the subject, I would be most grateful!

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    $\begingroup$ First, "This statement is false" is not a mathematical statement. But statements that are like that statement can be formulated. $\endgroup$ Sep 11, 2013 at 20:14
  • $\begingroup$ Pretty close to math.stackexchange.com/questions/69353/true-vs-provable except we are talking about false-vs-unprovable. $\endgroup$
    – rschwieb
    Sep 11, 2013 at 20:16
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    $\begingroup$ Could you please elaborate Thomas? I'm not exactly sure what you mean by that. Do you mean mathematical statements can encompass the same 'flaws' behind "this statement is false"? $\endgroup$
    – user93826
    Sep 11, 2013 at 20:17

2 Answers 2


Unprovable means that we cannot prove it. Given a list of axioms, $\varphi$ is provable from these axioms if there is a finite sequence of statements all of which are axioms, or deduced from axioms and previously written statements in the sequence, such that the final statement in the sequence is $\varphi$.

So when is $\varphi$ unprovable? When such finite sequence doesn't exist. How do we prove that something doesn't exist? In particular something like that?

Luckily we know that if $\sf ZF$ proves $\varphi$ then in every model of $\sf ZF$ the statement $\varphi$ must be true, on the other hand if we can only show there is some models of $\sf ZF$ where $\varphi$ is true then $\varphi$ is consistent with $\sf ZF$, being provable implies being consistent with, but usually not vice versa (in particular in the case of $\sf ZF$). So if we can find two models of $\sf ZF$, one where $\varphi$ is true and another where it's not, then we show that $\varphi$ is not provable from $\sf ZF$.

This is how the continuum hypothesis was proved to be unprovable. If $\sf ZF$ is consistent (otherwise it proves everything anyway), then it has a model where $\sf CH$ is true, this was shown by Goedel in the 1940's. Some 20 years later Cohen showed how to construct from Goedel's model a model of $\sf ZF$ where the continuum hypothesis is false. This construction is technical and requires more than basic understanding of set theory, so I won't get into it right now.

But all combined we have that if $\sf ZF$ is consistent to begin with, then it is consistent with the continuum hypothesis and with its negation. Meaning that the continuum hypothesis is unprovable from $\sf ZF$.


The unprovability of Truth and Falsity of CH within ZFC system means:

Under the consistence assumption of ZFC system, if you add CH or its negation to ZFC, the system remains consistent, and therefore CH or its nagation cannot be deduced from ZFC.


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