# Sequence of discontinuous functions

• rapple
In summary, the conversation discusses the need for an example of a sequence of functions that is discontinuous at every point on [0,1] but converges uniformly to a function that is continuous at every point. The dirichlet's function is suggested as a template, where f_n(x) = 1/n if x is rational and 0 if x is irrational, but the analysis is deemed erroneous because 1/n eventually goes to 0 and f_n(x) will be continuous as n->infinity. The conversation then asks for an example of a value of n where f_n is continuous, and it is clarified that while the limit is 0, f_n is not equal to zero for any finite n.

## Homework Statement

Need an example of a sequence of functions that is discountinuous at every point on [0,1] but converges uniformly to a function that is continuous at every point

## The Attempt at a Solution

I used the dirichlet's function as the template
f_n(x) = 1/n if x is rational and 0 if x is irrational

f_n(x) is discontinuous at every x in [0,1] and converges to f(x)=0

But this seems to be a erroneous analysis, because 1/n eventually goes to 0 so f_n(x) will be continuous as n->infinity

Can i get help in constructing this?

You already have a good example. What do you mean "because 1/n eventually goes to 0 so f_n(x) will be continuous as n->infinity". Can you give me an example of a value of n where f_n is continuous?

Dick said:
Can you give me an example of a value of n where f_n is continuous?

Since lim n->inf (1/n)=0, as n-> infinity, f_n(x) will be 0 for rationals as well.

which means that for any epsilon>0, if n is large enough, |f(x)-0|< epsilon for rational as well?

Less than epsilon, yes. Equal to zero, no. No f_n is equal to zero. Just because the limit is 0, that doesn't mean f_n becomes zero for any finite n.