Show that $C(\mathbb R, \mathbb R)$ is not connected

Question

$\bar{d}: \mathbb R × \mathbb R → \mathbb R$ denotes the standard bounded metric on $\mathbb R$ defined by $\bar{d}(x, y) := \min\{|y − x|, 1\}$.

Given a topological space $X$, let $C(X, \mathbb R)$ denote the set of all continuous functions $f : X → \mathbb R$ topologized with the topology of the uniform metric $ρ(f, g) :=\sup\{\bar{d}(f(x),g(x)) :x \in X\}$ for $f, g \in C(X, \mathbb R)$

Show that $C(\mathbb R, \mathbb R)$ is not connected.

Hint: let $p \in C(X,\mathbb R)$ be bounded and let $q \in C(X, \mathbb R)$ be unbounded. Are $p, q$ in the same connect led component?

My attempt:

$p$ is bounded, then there exists $M \in \mathbb R$ such that $|p(x)| \leq M$.

$q$ is unbounded, then $|q(x)| < inf$.

$ρ(p, q) :=\sup\{\bar{d}(p(x),q(x)) :x \in X\} = \sup\{\min\{|q(x)-p(x)|,1\}:x \in X\}$ for $p, q \in C(X, \mathbb R)$

Now $|q(x) - p(x)| \leq |q(x)| + |p(x)| < inf + M = inf$

So, $\min\{|q(x) - p(x)|, 1\} = 1$, and $\sup\{\min\{1\}\}= 1$. Does this tell me that $p, q$ are in different components, and so $C(\mathbb R, \mathbb R)$ is not connected?

Answer 1

Let $P,Q$ be the subspaces respectively of bounded and unbounded continuous functions on $\mathbb{R}$.

1. $P$ and $Q$ are disjoint and $P\cup Q=\mathcal{C}(\mathbb{R},\mathbb{R})$: a function cannot be both bounded and unbounded, but it must be one of the two
2. Every $p\in P$ satisfies $\overline{d}(p,Q)=1$, because necessarily for all $q\in Q$, $\sup (|p-q|)=+\infty$, and thus $\overline{d}(p,q)=1$. This implies that $P$ is open, since every ball of radius less than $1$ is totally contained in $P$. The same result holds for $Q$ by a similar argument

Thus, the space is not connected, since it is the disjoint union of open sets.

Note that one can prove (as hinted by TheSilverDoe in the comments to the OP, and proved here), that $P$ is path connected.