$\def\bbR{\mathbb R}\def\ssp{\kern.4mm}$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 $[\ssp-n,n\,]$ .



Is the strong Whitney topology connected?
No. It is not a connected space. We can in fact describe the connected components of this space quite easily. Let $\simeq$ be the equivalence relation on $C$ where $f\simeq g$ iff $f-g$ has compact support. I claim that the equivalence classes of $\simeq$ are precisely the components. Incidentally, the equivalence classes of $\simeq$ are the path components and quasi-components as well.


If $f-g$ has compact support, then let $L:[0,1]\rightarrow C$ be the mapping where $L(t)=g\cdot t+f\cdot(1-t)$ whenever $t\in[0,1]$. Then clearly $L$ is a path from $f$ to $g$. 


Now assume that $f\not\simeq g$. Then $f-g$ does not have compact support. Therefore, without loss of generality, assume that there are arbitrarily large positive real numbers $x$ such that $(f-g)(x)\neq 0$. Then there is some increasing sequence $(x\_{n})\_{n\in\omega}$ of real numbers with $x\_{n}\rightarrow\infty$ where $(f-g)(x\_{n})\neq 0$ for all $n$. Therefore let $\equiv$ be the equivalence relation on $C$ where $h\_{1}\equiv h\_{2}$ if and only if 
$$\lim\_{n\rightarrow\infty}\frac{(h\_{1}-h\_{2})(x\_{n})}{(f-g)(x\_{n})}\rightarrow 0.$$


Clearly $\equiv$ is an equivalence relation. On the other hand, the equivalence relation $\equiv$ partitions $C$ into open sets. Suppose that $U$ is an equivalence class in $C$ and $h\in U$. Then let $u$ be a continuous function with $u(x\_{n})=\frac{1}{n}\cdot|(f-g)(x\_{n})|$. Then if $|(k-h)(x)|<u(x)$ for all $x\in\mathbb{R}$, then $$|\frac{(k-h)(x\_{n})}{(f-g)(x\_{n})}|\leq|\frac{u(x\_{n})}{(f-g)(x\_{n})}|=\frac{1}{n}\cdot|\frac{(f-g)(x\_{n})}{(f-g)(x\_{n})}|=\frac{1}{n}$$, so 
$$\lim\_{n\rightarrow\infty}\frac{(k-h)(x\_{n})}{(f-g)(x\_{n})}\rightarrow 0.$$ 


Therefore $k\equiv h$, so $k\in U$ as well. Therefore $U$ is an open set. However, we have $f\not\equiv g$. Therefore $f$ and $g$ belong to different quasi-components, so $f$ and $g$ belong to different components.


Therefore $f,g$ belong to the same component if and only if $f\simeq g$.


This proof is the same proof as the proof of the fact that two sequences in the box topology on $\mathbb{R}^{\omega}$ are in the same component if and only if they agree except for possibly finitely many coordinates.


