If $f$ is continuous and $f_n \rightarrow f$ uniformly on $X$ as $n\rightarrow \infty$

Question

Let $(X, d_X)$ and $(Y,d_Y)$ be metric spaces, let $f : X \rightarrow Y $, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n : X \rightarrow Y$ .

Prove or give a counterexample to the following statement: If $f$ is continuous and $f_n \rightarrow f$ uniformly on $X$ as $n\rightarrow \infty$, then there must exist some $N \geq1$ such that $f_n$ is continuous for each $n\geq N$.

I believe that the following statement is not true, and my counterexample lies within the following lines.

I know that it has to be something with $\frac{1}{n}$ but I know we hvae to modify this a bit in order to account for the continuity, I would really appreciate some help in modifying my counterexample so it fully works, thank you!

Answer 1

Since you are looking for an counterexample, you can simplify the setting. For example, let be $X=Y=\mathbb R$ and $d_X$ and $d_Y$ the euclidean metric.

Now, you can choose $f$ to be constant, for example $f\equiv 0$. Next, you can construction a sequence of function $(f_n)_{n=1}^\infty$ with an jump, which becomes $0$ for $n\to\infty$. Since each $f_n$ has a jump, they are all not continuous, while $f$ is continouous.

One simple example for a sequence $(f_n)_{n=1}^\infty$ with $f_n\to f$ uniformly.

Define $$f_n(x)=\begin{cases}\frac1n & x>0\\0&x\leq 0\end{cases}$$

Or a bit more extreme example for a sequence $(f_n)_{n=1}^\infty$ with $f_n\to f$ uniformly.

Define $$f_n(x)=\begin{cases}\frac1n & x\in\mathbb Q\\0&x\in\mathbb R\setminus\mathbb Q\end{cases}$$ In this example, $f_n$ is discontinuous everywhere for all $n\in\mathbb N$, while $f$ is continuous.