Prove that the sequence $(f_n)_{n \geq 1}$ does not have a convergent subsequence

Question

Given the norm

$$ \| f \| := \sup \{ |f(x)| \mid x \in \mathbb{R} \}$$ Let $V$ be the set of bounded functions $f : \mathbb{R} \rightarrow \mathbb{R}$.

(1) Show that for every $f \in V$ the norm exists.

For this question I started with:

Let $f \in V$, then $f$ is bounded, so there exists an $R > 0$ with $$|f(x)| < R$$ for all $x$. This means $R$ is an upper bound of the set $\{|f(x)| \mid x \in \mathbb{R} \}$. Now I think I need to show that the set of upper bounds of that set has a minimum, but I am not sure how to do this.

Now let the set $B$ be given with $$B = \{ f \in V \mid \| f \| \leq 1 \}. $$ For every $n \in \mathbb{N}$ de function $f_n : \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f_n = 1$ on $[2n, 2n + 1]$ and $f_n = 0$ on $\mathbb{R} \setminus[2n, 2n + 1]$.

(2) Show that the sequence $(f_n)_{n \geq 1}$ is in set $B$ and does not have a convergent subsequence.

For this question I couldn't find anything yet. Maybe I can use the fact that set $B$ is closed and bounded?

Answer 1

For $(1)$, a minimum upper bound exists by the supremum axiom/condition: every set that is bounded above has a least upper bound, i.e. a supremum. This is a mainstay of real analysis.

For $(2)$, show that $\|f_n - f_m\| = 1$ if $n \neq m$. This prevents there being a Cauchy subsequence, let alone a convergent subsequence; for $\varepsilon = 1/2$, there cannot be any $N$ such that $m, n > N \implies \|f_n - f_m\| < \varepsilon$, since choosing any $m > n > N$ will make $\|f_n - f_m\| = 1$.

Answer 2

If $x\in\mathbb R$, then $f_n(x)=0$ if $n$ is large enough. Therefore, if a subsequence of the sequence $(f_n)_{n\in\mathbb N}$ converges, it can only converge to the null function. But $(\forall n\in\mathbb N):\lVert f_n-0\rVert=1$.