Assume supf(X)> supg(X)........................................... ................................................1

From the definition of the supremum we have:

for all x , if xεΧ then .................................................. ...........................................2

for all x, if xεΧ then .................................................. ............................................3

AND

for all ε>0 and hence for ε = supf(X)-supg(X)>0 there exists a yεf(X) such that :

supf(X)-(supf(X)-supg(X)< y or

supg(X) <y .................................................. ...........................................4

But since y belongs to f(X) y= f(x) and xεΧ and since for all xεΧ , (4) becomes:

supg(X)< g(x).............................................. ........................................5

And using (3) we end up with the contradiction:

supg(X) < supg(X)

the proof for the infemums is a similar one