# Please clarify proof that $\mathbb R_K$ is not path connected.

Topology by James Munkres Exer27.3c

Here is dbfin's pf:

I am really confused because of all the $$\mathbb R$$ vs $$\mathbb R_K$$. I have found other proofs that I understand but would like to understand this one.

Is my understanding of this proof right? Also, please fill in gaps for where I am really not sure, mostly 4&7 below.

1. Suppose on the contrary $$\mathbb R_K$$ is path connected. By definition, for any pair of points $$x,y$$ in $$\mathbb R_K$$, we can construct a path in $$\mathbb R_K$$ from $$x$$ to $$y$$, i.e. a continuous a function $$f:[a,b] \to \mathbb R_K$$ s.t. $$f(a)=x$$ and $$f(b)=y$$, where $$[a,b]$$ is a subspace of $$\mathbb R$$. Choose $$x=0$$ and $$y=1$$.

2. $$f([a,b])$$ is compact and connected as a subspace of $$\mathbb R_K$$.

• This is because $$[a,b]$$ is a compact and connected subspace of $$\mathbb R$$, and images of compact and connected subspaces of the domain of a continuous function are compact and connected subspaces of the range of the function.
3. $$f([a,b])$$ is connected in $$\mathbb R$$.

• This is because the topology of $$\mathbb R_K$$ is finer than the standard topology of $$\mathbb R$$.
4. $$f([a,b])$$ is convex in $$\mathbb R$$.

• I am really not sure why. I guess this is because for any set $$A \subseteq \mathbb R$$, $$A$$ is connected iff $$A$$ is an interval or ray iff $$A$$ is convex. I don't think all of these were proven at this point in the book. What has been proven so far I believe is that:

• Any interval or ray in an ordered set X is convex in X and

• Any interval or ray in $$\mathbb R$$ is connected.

• I don't think it has been proven that the only connected sets of $$\mathbb R$$ are intervals or rays. I was actually expecting convex implies connected.

5. $$[0,1] \subseteq f([a,b]) \subseteq \mathbb R$$

• One might argue Intermediate Value Theorem (Thm 24.3), but here, this is because of (4), I guess.
6. $$[0,1]$$ is a closed subspace of $$f([a,b]) \subseteq \mathbb R$$.

• This is because compact subspaces of Hausdorff spaces are closed and subspaces of Hausdorff spaces are Hausdorff.
7. $$[0,1]$$ is a compact subspace of $$\mathbb R_K$$.

• 7.1. (6) implies that $$[0,1]$$ is a closed subspace of $$f([a,b]) \subseteq \mathbb R_K$$ because the topology of $$\mathbb R_K$$ is finer than the standard topology of $$\mathbb R$$. --> Is this right?!

• 7.2. This implies that since $$f([a,b])$$ is a compact subspace of $$\mathbb R_K$$, we have that $$[0,1]$$ is a compact subspace of $$f([a,b]) \subseteq \mathbb R_K$$ because closed subspaces of compact subspaces are compact subspaces.

• 7.3. This implies (7), by Compact subspaces of compact subspaces are compact subsubspaces.

Consider the interval $$[c,d]\supseteq [0,1]$$, then $$[c,d]=([c,d]-K)\bigcup (\cup_{n=1}^{\infty} (\frac{1}{n+1},1])$$ has no finite subcovering( otherwise we have $$m\in \Bbb N$$ such that $$[c,d]=([c,d]-K)\bigcup (\cup_{n=1}^m (\frac{1}{n+1},1])$$ which is impossible) . Therefore $$[c,d]$$ is not compact in $$\Bbb R_K$$.

If possible let $$\Bbb R_K$$ is path connected then the points $$0,1$$ can joined by some path $$f:[a,b]\rightarrow \Bbb R_K$$ such that $$f(a)=0,f(b)=1$$. But continuous image of a connected set is connected. Since $$[a,b]$$ is connected w.r.t. subspace topology inherited from standard topology on $$\Bbb R$$. $$image(f)$$ is connected set in $$\Bbb R_K$$. Then $$image(f)$$ is also connected is $$\Bbb R_S:= \Bbb R \ with\ standard\ topology$$. ( otherwise $$image(f)$$ can be written as two disjoint union of two non-empty open sets of $$\Bbb R_S$$ ,but open sets of $$\Bbb R_S$$ are also open in $$\Bbb R_K$$ hence $$image(f)$$ can also be written as two disjoint union of non-empty open sets in $$\Bbb R_K$$, contradictoriy to the assumption that $$image(f)$$ is connected in $$\Bbb R_K$$). Now connected subsets of $$\Bbb R_S$$ are intevals i.e. $$image(f)$$ is an interval containing $$0,1$$. Therefore $$image(f)\supseteq [0,1]$$.

Also $$[a,b]$$ is compact w.r.t. subspace topology inherited from $$\Bbb R_S$$ and since continuous image of compact set is compact we have $$image(f)$$ is compact w.r.t. subspace topology inherited from $$\Bbb R_K$$. Therefore $$image(f)$$ is also compact w.r.t. subspace topology inherited from $$\Bbb R_S$$ (otherwise $$image(f)\subseteq \cup_{i\in \mathscr A} U_i$$ for some index set $$\mathscr A$$ and some open sets $$U_i$$ in $$\Bbb R_S$$ such that this cover has no finite subcover, since open set in $$\Bbb R_S$$ is also open in $$\Bbb R_K$$ , we can say also that $$image(f)$$ is also not compact in $$\Bbb R_K$$, contradiction). Therefore $$image(f)$$ is an compact interval of the form $$[c,d]$$ containing $$[0,1]$$.

Now by 1st argument any interval of the form $$[c,d]$$ containing $$[0,1]$$ is not compact in $$\Bbb R_K$$.

• Thanks UserS. So, your argument indeed relies on connected in $\mathbb R_S$ implies interval (or ray) in $\mathbb R_S$? And this is the justification for #4 in dbfin's proof? – BCLC Oct 19 '18 at 11:49
• Also, how precisely did you arrive at the contradiction in the last part? Is the #7 above right? – BCLC Oct 19 '18 at 11:58
• I first show $image(f)$ is compact and connected w.r.t. subspace topology inherited from $\Bbb R_K$. Hence $image(f)$ is also compact and connected w.r.t. subspace topology inherited from $\Bbb R_S$.Now what are the connected compact subspaces of $\Bbb R_S$ , they are nothing but, intervals of the form $[c,d]$ where $c,d\in \Bbb R$. Finally from the 1st paragraph we can say any interval $[c,d]$ containing $[0,1]$ is non-compact w.r.t. subspace topology inherited from $\Bbb R_K$. But image of $f$ which is $[c,d]$ is compact due to continuity of $f$. – Sumanta Oct 19 '18 at 12:50
• That is we use standard topology to determine how does the image of $f$ look like and showed that it is of the form [c,d] containing [0,1] but this type of intervals are non compact w.r.t. K-topology. Hence we arrive at a contradiction , since continuous image of compact set is compact. – Sumanta Oct 19 '18 at 12:54
• UserS, but why do you deduce $[c,d]$ compact in $\mathbb R_K$ rather than just $\mathbb R = \mathbb R_S$? I mean, what justifies #7 in the proof? – BCLC Oct 19 '18 at 12:55

To see that $$\mathbb{R}_K$$ is not path-connected, suppose there is a continuous path $$f:[0,1] \to \mathbb{R}_K$$ such that $$f(0)= 0$$ and $$f(1)=1$$. Then $$C:=f[[0,1]] \subseteq \mathbb{R}_K$$ is compact and connected in $$\mathbb{R}_K$$ because $$[0,1]$$ is compact and connected and these properties are preserved under continuous maps.

Now, as the topology on $$\mathbb{R}_K$$ is finer than the usual topology on $$\mathbb{R}$$, $$C$$ is also usual-topology compact and connected (a usual topology disconnection for $$C$$ would also be one for $$\mathbb{R}_K$$, and a usual open cover is also one for $$\mathbb{R}_K$$.

So $$C$$ is a usual compact-connected set that contains $$0$$ and $$1$$ so $$[0,1] \subseteq C$$. But $$K= \{\frac{1}{n}: n \ge 1\} \subseteq C$$ as well. And $$K$$ is closed and discrete in $$\mathbb{R}_K$$ so also in $$C$$. This contradicts the compactness of $$C$$ ($$K$$ would then also be compact but as infinite discrete space this is impossible).

This final contradiction shows that $$f$$ cannot exist and $$\mathbb{R}_K$$ is not path-connected.

Another contradiction could be achieved noting that $$C$$ would be non-regular (adapt the non-regularity proof for $$\mathbb{R}_K$$) and compact Hausdorff spaces are normal and regular.

On dbfin.com I am not doing your homework for you, often leaving the proofs without all details that say your grader would require. So there is always some space for you to think and understand what is going on. And you should do it yourself, not asking on other websites. But in this case you are fine as you show all your hard work trying to understand each step in my logic.

I think you have nailed it, though it seems nobody has addressed your concerns. The other two answers simply paraphrase the proof.

1. A hint: Theorem 24.3. Yes, it is the intermediate value theorem stated for a function from a connected to an ordered set (btw, this is why you cannot use it for $$\mathbb{R}_K$$). Now think how to directly from this theorem obtain the statement that if a subset of $$\mathbb{R}$$ is connected then it is convex.

2. Yes, (4) and NOT the IVT, see above. This is basically why I do (4). In essence, I prove the IVT for $$\mathbb{R}_K$$ using the fact that it is finer than the ordered $$\mathbb{R}$$.

3. $$[0,1]$$ is the complement of the union of two open (in both topologies) intervals.

4. 7.1. $$[0,1]$$ is closed in $$\mathbb{R}_K$$, see above. 7.2-7.3. Right, except you have a typo: closed subsets of compact spaces are compact.