Let the sequence of functions $f_n(x)$ equal $1$ if $x ∈ [n, n + 1)$ and $0$ otherwise. Why doesn't $f_n$ converge uniformly?

Question

Let the sequence of functions $f_n(x)$ equal $1$ if $x ∈ [n, n + 1)$ and $0$ otherwise. How can I use the definition of uniform convergence to show that $f_n$ does not converges uniformly?

If a function is uniformly convergent, then for all $\epsilon > 0$ there exists an integer $N$ such that for all $n \geq N$ adn for all $x$, $|f_n(x) - f(x)| < \epsilon$. Therefore we wish to find some $\epsilon > 0$ such that for each $n$ there is an $x ∈ [n, n + 1)$ such that $|f_n(x) - f(x)| \geq \epsilon$. But how can I do this ?

Answer 1

Notice that for every $x$, $f_n(x)\to 0$ as $n \to \infty$, or in other words, $0$ is the pointwise limit of $f_n$. Then if $f_n$ converges uniformly to some function $f$ this must be the constant function $0$.

Now let $\epsilon = \frac 12$ and notice that the convergence is not uniform since for every $n$, if $x \in [n,n+1)$ then $$|f_n(x) - f(x)| = 1 > \frac 12 = \epsilon.$$

If you wish, for every $n$, $x_n = n + \frac 12$ does the trick.