Calculating number of equivalence classes where two points are equivalent if they can be joined by a continuous path.
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
- Only one.
- At most finite.
- At most countable.
- 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.
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.
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.