# Two questions on functions

• Nov 12th 2009, 01:12 PM
jmoney90
Two questions on functions
Okay, here is one:

Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha

So far, I know that you're going to have to make use of the definition of continuity. THat's about as far as I got though.

Another:

Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x
• Nov 12th 2009, 01:23 PM
Plato
Quote:

Originally Posted by jmoney90
Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha

In the definition of continunity let $\displaystyle \varepsilon = \frac{{f(b) - \alpha }}{2} > 0$.

Now the way this forum works, you should show some effort on your part.
• Nov 12th 2009, 01:40 PM
hjortur
Think a bit about the last one, what happens if $\displaystyle f(x)=k>0$ where x is irrational?
What if you pick $\displaystyle \varepsilon$ such that $\displaystyle 0<\varepsilon<k$?

Know that every open interval contains irrational and rational numbers.

Try to visualize it.
• Nov 12th 2009, 06:10 PM
jmoney90
Quote:

Originally Posted by Plato
In the definition of continunity let $\displaystyle \varepsilon = \frac{{f(b) - \alpha }}{2} > 0$.

Now the way this forum works, you should show some effort on your part.

I'm still not seeing what I should be doing. I substituted it into the definition and tried tinkering around, but I'm not seeing anything.
• Nov 13th 2009, 12:39 AM
hjortur
Try this:
Define a function $\displaystyle g(x)=f(x)-\alpha$. g is continuous, because f(x) is and $\displaystyle \alpha$ is a constant.

Now $\displaystyle g(b)>0$, and if g is continuous then there must be an interval $\displaystyle I$ about b such that for every $\displaystyle x\in I$ , $\displaystyle g(x)>0$.

Is it easier to see now ?
• Nov 13th 2009, 05:05 AM
HallsofIvy
Quote:

Originally Posted by hjortur
Think a bit about the last one, what happens if $\displaystyle f(x)=k>0$ where x is irrational?
What if you pick $\displaystyle \varepsilon$ such that $\displaystyle 0<\varepsilon<k$?

Know that every open interval contains irrational and rational numbers.

Try to visualize it.

I don't see how being rational or irrational has anything to do with this problem!
• Nov 13th 2009, 05:32 AM
hjortur
Quote:

Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x
Then please let me know how you solve that without referring to rational or irrational numbers.