$\def\bbR{\mathbb R}\def\ssp{\kern.4mm}$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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.