Is it possible to construct a "divergent" sequence of functions?

Question

I'm trying to understand pointwise versus uniform convergence of functions. So I want to find a sequence $\{f_n\}$ of functions $[0,1] \to \mathbb{R}$ that fails to converge pointwise to a function from $[0,1] \to \mathbb{R}$. I'm not 100% sure whether this is possible. Because $[0,1]$ is compact and therefore complete, it isn't possible to build a sequence of real numbers that converges to something outside of $[0,1]$, so I don't think I can build a sequence $\{f_n\}$ which converges to a function $f$ with domain $S \supset [0,1]$. So I think to find an example I would need a sequence that doesn't pointwise converge, which I think needs to happen only for a particular $x$. I can try to come up with a sequence of discontinuous functions $\{f_n\}$ which would break pointwise convergence at a single point, but every example I've tried so far hasn't worked.

Answer 1

You are right about the part that a convergent sequence can not converge to something outside the interval $[0,1]$. However, the sequence does not have to converge at all. For example, $x_n=(-1)^n$ is in $[-1,1]$ but does not converge. Having this in mind, you could define your $f_n(x)=(-1)^n,\ \forall x\in [0,1]$. This way, your function sequence does not converge pointwise.