Show that there exists no sequence of functions satisfying the following

[poet_3000]

I found this interesting exercise on a topology book I'm reading, but I don't have a clue what to do.

Show that there is no sequence {g_n} of continuous functions from R to R such that the sequence {(g_n)(x)} is bounded iff x is rational (where R = set of real numbers).

 

[lavinia]

I found this interesting exercise on a topology book I'm reading, but I don't have a clue what to do.

Show that there is no sequence {g_n} of continuous functions from R to R such that the sequence {(g_n)(x)} is bounded iff x is rational (where R = set of real numbers).

Every real is the limit of a sequence of rational. The value of a continuous function on this sequence of rationals converges to its value on the limit.

 

[Tinyboss]

Define $$B_n=\{x\in\mathbb{R}:|f_k(x)|>n\:\mathrm{ for some }\:k\}$$ and note that each $$B_n$$ is open. Now the set of points where the sequence is unbounded is $$\bigcap_{n\ge0}B_n$$. This is a $$G_\delta$$ set, and the rationals are not a $$G_\delta$$ set.

http://en.wikipedia.org/wiki/Gδ_set

 

[ForMyThunder]

Tinyboss,

You are a clever, clever man. How did you think of something like that?