# Thread: infimums and supremums of Functions

1. ## infimums and supremums of Functions

Let f and g be bounded functions from a nonempty set X into R.

Prove that if f(x) <= g(x) for all x in X, then inf f(X) <= inf g(X) and
sup f(X) <= sup g(X).

2. Originally Posted by Jdg6057
Let f and g be bounded functions from a nonempty set X into R.

Prove that if f(x) <= g(x) for all x in X, then inf f(X) <= inf g(X) and
sup f(X) <= sup g(X).
Assume supf(X)> supg(X)........................................... ................................................1

From the definition of the supremum we have:

for all x , if xεΧ then $f(x)\leq supf(X)$.................................................. ...........................................2

for all x, if xεΧ then $g(x)\leq supg(X)$.................................................. ............................................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 $\leq supf(X)$ or

supg(X) <y $\leq supf(X)$.................................................. ...........................................4

But since y belongs to f(X) y= f(x) and xεΧ and since for all xεΧ , $f(x)\leq g(x)$ (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