Sequences and functions homework help

[JasonJo]

let f,g be continuous functions from R to R and suppose that f(x)=g(x) for all rational points. prove that f(x)=g(x) for all x in R.
- i said that we know that since given any real number c, there exists a rational sequence (xn) such that xn converges to c, therefore we conclude that f(xn)=g(xn)=f(c)=g(c), and c is any real number, QED.
is this a good proof?

what about if we only knew that f(1/n)=g(1/n) for all n a natural number?
- i said we know that f(0)=g(0) since the sequence 1/n converges to 0, but other than that, not much else...am i missing something?

prove that for every c a real number, there exists a rational sequence converging to c.
- I am not sure how to prove this, my professor said to use the Nested Intervals Property, but I can't really establish the correct intervals

prove that for every c a real number, there exists an irrational sequence converging to c.
- again, not too sure about this one

 

[Office_Shredder]

Your proof is a bit sparse. It's a good start, but you should really drive home the full definition of a continuous function, and what it means for a sequence to converge to c. So you should say given an epsilon, we know there is some N such that all x_n with n>N means |x_n-epsilon|< x, and then use that with the definition of continuity.

EDIT: I note you're using a property in part 3, so you definitely should prove part 3 without using part 1 (I don't know how you would use part 1, but make sure you don't).

 

[JasonJo]

any help with the sequences part? i don't get how to apply Nested Intervals Property

 

[quasar987]

One way is to consider an class of intervals $I_n=(c-1/n,c+1/n)$. Then the infinite intersection of the $I_n$s is c (is this what he calls the nested interval property?)

Now construct a sequence by taking (any) one rational number in each $I_n$ (there is one since the rationals are dense in R). Show that this sequence will converge to c.

Of course the same thing works by taking an irrational number in each $I_n$.