*Originally posted by mtthw*

**Well, yes, I'm sure a full proof would lead to several pages and would contain a lot of maths. I've no inclination to try and produce that, and I don't suppose anyone want to read it if I did. These are always going to be sketch proofs, with the assumption that they are built upon other proofs. It only really matters if someone wants to start disputing one of those foundations.**

indeed.

the proof is relatively simple if the isobar forms a closed interval [a,b]. We can simply invoke the Extreme Value Theorem of continuous functions combined with the fact that for the continuous temperature function f f(a) = f(b)

Now let f(x) = K be the highest value of f. (if x=a then let f(x) be the lowest value of f and reverse the signs, if x is still a then we have f(x) = K for all x and QED).

Now pick a J < K. and look at the interval [a,x] according to the intermediate value theorem there must exist a point x' in [a,x] for which f(x'ðŸ˜‰ = J. Equally when we look at the interval [x,b] there must exist a point x'' in [x,b] for which f(x''ðŸ˜‰ = J

Therefore we have f(x'ðŸ˜‰ = f(x''ðŸ˜‰ for some x' /= x'' and QED

The proof is a little more complex if the isobar is a half open interval [a,b) (it is still true we just need to take a closed subset of [a,b) call it [a,c] and show that if there doesn't exist a x' /= x'' for which f(x'ðŸ˜‰ = f{x''ðŸ˜‰ then we can always pick another c -> b which does)