real analysis - If $f$ is continuous and $f_n rightarrow f$ uniformly on $X$ as $nrightarrow infty$

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

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.