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

  • $\begingroup$ No. It is not conected. $\endgroup$ – Joseph Van Name Feb 8 '15 at 22:06
  • $\begingroup$ @Joseph Van Name: Can you give a proof, hint or reference? $\endgroup$ – TaQ Feb 8 '15 at 22:17
  • $\begingroup$ After leaving my above comment I closed my computer but almost immediately after that I found a simple example of sets $U$ and $V$ showing that $X$ is not connected. I came back having as my intention to give it as an answer but I saw that Joseph Van Name had just started writing his answer but had interrupted for some reason. I give below my own answer but I accept Joseph's since he succeeded to give of a complete description the connected components of $\mathscr T$ . $\endgroup$ – TaQ Feb 9 '15 at 7:38

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.

  • $\begingroup$ Joseph, you need to fix the typos of the type $\ f-g(x_n);\ $ it should be $\ (f-g)(x_n)\ $. and similar on similar occasions. $\endgroup$ – Włodzimierz Holsztyński Feb 9 '15 at 2:18
  • $\begingroup$ I added the parentheses. I guess I was too concerned about conserving parentheses. $\endgroup$ – Joseph Van Name Feb 9 '15 at 2:32
  • $\begingroup$ You have a complete answer--perfect. $\endgroup$ – Włodzimierz Holsztyński Feb 9 '15 at 2:46

$\def\ssp{\kern.4mm}$Take any $f\in C$ and let $U$ be the set of all $g\in C$ with $\lim_{\,t\to+\infty\,}(f-g)(t)=0$ and put $V=C\setminus U$ . Then $U,V\in\mathscr T\setminus\{\ssp\emptyset\ssp\}$ and trivially $U\cup V=C$ and $U\cap V=\emptyset$ . Hence the space $X$ is not connected.


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