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

http://en.wikipedia.org/wiki/Gδ_set

 

[ForMyThunder]

Tinyboss,

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

 

[blue_raver22]

I would approach this problem by first understanding the definitions and concepts involved. In this case, we are dealing with sequences of functions, continuity, boundedness, and rational numbers.

To show that there exists no sequence of functions satisfying the given condition, we need to prove that it is not possible to find a sequence of functions that meets the given criteria. In other words, we need to show that there is a logical contradiction in the statement.

One approach to proving this is through proof by contradiction. Suppose there exists a sequence of continuous functions {g_n} from R to R such that {(g_n)(x)} is bounded if and only if x is rational. This means that for any rational number x, the sequence {(g_n)(x)} is bounded, and for any irrational number x, the sequence is unbounded.

However, this leads to a contradiction since the set of rational numbers and the set of irrational numbers are both dense in R. This means that between any two real numbers, there exists both a rational and an irrational number. Therefore, for any real number x, the sequence {(g_n)(x)} must be both bounded and unbounded, which is impossible.

Thus, we have shown that our initial assumption of the existence of such a sequence leads to a contradiction. Therefore, there exists no sequence of functions {g_n} that satisfies the given condition. This proves the statement and completes the exercise.