Uniform Continuity of a Piecewise Function

Question

Suppose you have the function $f : [0,1] \rightarrow \mathbb{R}$, with $f(x) = 0$ if $x \in [0,1)$ and $f(x)=1$ if $x=1$. Prove that it is uniformly continuous.

I got this function as the pointwise limit of $f_{n}(x)=x^{n}$ and was wondering if it was uniformly continuous since each of the $f_{n}(x)$ are uniformly continuous. If so, how?

Answer 1

Uniform continuity implies continuity. So, since your function is discontinuous at $1$, it is not uniformly continuous.

Answer 2

The function is not uniformly continuous. It is not even continuous. ($1$ is a point of discontinuity.)

Answer 3

It cannot be uniformly continuous, since that would imply that $f$ is continuous, while it is clearly not the case.

In general, when $f_n \to f$ pointwise, you cannot expect that the limit is continuous, even if every $f_n$ is.

In order to guarantee that $f$ is continuous, you need the sequence $f_n$ to converge uniformly to $f$.