$\def\bbR{\mathbb R}\def\ssp{\kern.4mm}$Put more precisely, let $C$ be the set of all continuous functions $f:\bbR\to\bbR$ when
$\bbR$ has its standard order topology. Let $\mathscr T$ be the set of all $U\subseteq C$ with the property that for every $f\in U$ there exists a continuous function $u:\bbR\to\bbR^+=\bbR\cap\{\,t:t>0\,\}$ such that we have $g\in U$ whenever $g\in C$ is such that $|\,f(t)-g(t)\,|<u(t)$ holds for all $t\in\bbR$. Then $X=(C,\mathscr T)$ is a Hausdorff topological space, and it is relatively easy to see that in $X$ no point has any pathwise connected neighbourhood, cf. for example Lemma 41.7 on page 435 in Kriegl and Michor's book *The Convenient Setting of Global Analysis*. So $X$ is not locally pathwise connected and not pathwise connected.

Question.Is $X$ a connected topological space?

That is, do there exist $U,V\in \mathscr T\setminus\{\ssp\emptyset\ssp\}$ with $C=U\cup V$ and $U\cap V=\emptyset$ ? I have not been succesfull neither in finding a proof nor a counterexample.

Observe that if we construct a stronger topology $\mathscr T_1$ by modifying the above definition so that it is also required that the set $\bbR\cap\{\,t:f(t)\not=g(t)\,\}$ be relatively compact, then $\mathscr T_1$ *is* locally pathwise connected and *is not* connected. This topology $\mathscr T_1$ is the manifold topology for an infinite-dimensional smooth (affine) manifold structure for $C$ modelled on the space ${\rm ind\,lim\,}_{\,n\to+\infty\,}F_n$ where $F_n$ is the subspace of the Banach space $C_0(\bbR)$ formed by the functions having support included in the closed interval $[\ssp-n,n\,]$ .