Q. Let $G$ be an open set in $\Bbb R^n$. Two points $x,y \in G$ are said to be equivalent if they can be joined by a continuous path completely lying inside $G$. Number of equivalence classes is

  1. Only one.
  2. At most finite.
  3. At most countable.
  4. Can be finite, countable or uncountable.

This question was asked in the NET exam December 2016.

We can discard the first option by taking $n=1$ and $G=(-\infty,0) \cup (0,\infty)$.

We can reject the second option by taking $n=1$ and $G=\cup_{k \in \Bbb Z} (k,k+1).$

Now fun begins. Can we get an uncountable number of disjoint open path connected subsets of $\Bbb R^n$ for some $n$? If so, then we can take $G$ to be their union. For $n=1$, this method fails because that would give us the contradiction that the set of irrational numbers is countable.

  • $\begingroup$ I don't follow what you're trying to say in the case $n=1$ at all... $\endgroup$ – Eric Wofsey Dec 20 '16 at 6:15
  • $\begingroup$ (3) follows from math.stackexchange.com/questions/640491/… $\endgroup$ – Henricus V. Dec 20 '16 at 6:17
  • $\begingroup$ @EricWofsey which statement about $n=1$? I have mentioned three cases. First rejected option 1. Second rejected option 2. and third gave me contradiction. $\endgroup$ – Error 404 Dec 20 '16 at 6:20
  • $\begingroup$ Sorry, I mean in the final sentence. $\endgroup$ – Eric Wofsey Dec 20 '16 at 6:20
  • $\begingroup$ @EricWofsey It's ok. My argument in the final sentence was if we suppose $G=\cup_{i} (q_i,q_{i+1})$ where $q_i \in \Bbb R - \Bbb Q$, then that would mean $q_i$s are countable. I am feeling like dumb now. :/ $\endgroup$ – Error 404 Dec 20 '16 at 6:28

It is impossible to have uncountably many equivalence classes. Note that each equivalence class is an open set, since balls are path-connected and so if $x\in G$ then any open ball around $x$ contained in $G$ is in the same equivalence class. Now any nonempty open subset of $\mathbb{R}^n$ contains an element of $\mathbb{Q}^n$, so each equivalence class must contain some element of $\mathbb{Q}^n$. Since $\mathbb{Q}^n$ is countable, there can be only countably many equivalence classes.

More generally, this argument applies with $\mathbb{R}^n$ replaced by any locally path-connected separable space.

  • $\begingroup$ Thanks for mentioning the general case of 'locally path-connected separable space.' $\endgroup$ – Error 404 Dec 20 '16 at 6:30
  • $\begingroup$ This proof seems incorrect to me. Suppose $G=\mathbb{Q}\subset\mathbb{R}$. Then each arc-wise connected component of $G$ is of the form $\{ q \}$ where $q$ is a rational number. This set $\{ q \}$ is not open in $\mathbb{R}$, and it is not even (relatively) open in $G$. $\endgroup$ – Jeppe Stig Nielsen Dec 20 '16 at 9:57
  • $\begingroup$ @JeppeStigNielsen $G$ is given to be an open set. How can we take $G=\Bbb Q$ ? $\endgroup$ – Error 404 Dec 20 '16 at 10:13
  • $\begingroup$ Ah, I thought there would be something I had misunderstood. Thanks. $\endgroup$ – Jeppe Stig Nielsen Dec 20 '16 at 10:22
  • $\begingroup$ @JeppeStigNielsen No problems. $\endgroup$ – Error 404 Dec 20 '16 at 10:24

As an open subset of $\mathbb R^n$ is the union of at most a countable number of open balls (centered on points with rational coordinates and having a rational radius), response 3. is the right one.

  • $\begingroup$ @VikrantDesai The point here is that any open set can be represented as a union of balls with rational centers and radii. I'd consider it to be a nice exercise to show this if you've never seen a proof. $\endgroup$ – Wojowu Dec 20 '16 at 9:34
  • $\begingroup$ @Wojowu Okay. True that. I am also thinking same about writing down the proof for it. Will surely try. $\endgroup$ – Error 404 Dec 20 '16 at 9:48

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