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.
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$\begingroup$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.
$\endgroup$ 5 $\begingroup$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.
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