1. Continuous Function Question

Suppose a function fa,b) -> R is continuous and f(r) = 0 for each rational number r in (a,b). Show that f(x) = 0 for each x element (a,b).

So by the definition of continuous every sequence (xn) element dom(f) converges to x0 and limf(xn) = f(x0). Then, xn = 0 must converge to x0. Then, limf(0) = lim0 = 0 = f(x0).

I was wondering if this was correct logic and if there other xn sequences that could be used.

Thanks

2. Originally Posted by tbyou87
Suppose a function fa,b) -> R is continuous and f(r) = 0 for each rational number r in (a,b). Show that f(x) = 0 for each x element (a,b).
So by the definition of continuous every sequence (xn) element dom(f) converges to x0 and limf(xn) = f(x0). Then, xn = 0 must converge to x0. Then, limf(0) = lim0 = 0 = f(x0).
I was wondering if this was correct logic and if there other xn sequences that could be used.
Yes the logic is correct. Suppose that $\displaystyle a < x_0 < b\quad \& \quad f\left( {x_0 } \right) \ne 0$.

3. If you were doing it by contradiction, does this make sense:

Assume f(x0) not equal 0 and a<x0<b.
By definition of continuity, lim (xn) = x0 and lim f(xn) = f(x0).
Since a<r<b, lim (xn) = r and lim f(xn) = f(r) = 0.

Thanks

4. Once again you have the right idea. But you need more detail.
Suppose that $\displaystyle f(x_0 ) \ne 0$. Every real number is the limit of a sequence of rational numbers.
So there is a sequence of rational numbers in (a,b) and $\displaystyle \left( {r_n } \right) \to x_0$. But $\displaystyle \left( {\forall n} \right)\left[ {f\left( {r_n } \right) = 0} \right]$.
Is that possible if $\displaystyle f(x_0 ) \ne 0$ and $\displaystyle \left( {f\left( {r_n } \right)} \right) \to f\left( {x_0 } \right)$ by continuity?

5. No because continuity states that (f(rn)) -> f(x0) = f(r) = 0.

6. Originally Posted by tbyou87
No because continuity states that (f(rn)) -> f(x0) = f(r) = 0.
Now you understand the the whole idea.

7. Can either of you help me with my limits problem? Its posted a few threads below this one. I need answers by tonite, so i really need help! Thanks.