0

In Need a hint: show that a subset $E \subset \mathbb{R}$ with no limit points is at most countable, the OP constructs an interval $N_{r_x}(x)$ for each $x \in E$ such that $N_{r_x}(x) \cap E = \{x\}$. It is then hinted that the solution follows from the denseness of $\mathbb{Q}$ in $\mathbb{R}$.

I tried to construct a map from $E$ to $\mathbb{Q}$ sending $x$ onto a rational number $q_x$ with $q_x \in N_{r_x}(x)$ (which always exists, since we can always pick a rational number between $x -r_x$ and $x+r_x$. However, how can we argue that this map is injective? Or is this not what is intended by the hint?

Student
  • 4,438

2 Answers2

2

Your proof is valid.

Your map is injective because your open balls are pairwise disjoint.

If $x\ne y$ then your $q_x$ and $q_y$ are in two disjoint open balls centered at $x$ and $y$ so $q_x \ne q_y$, that is the set is disjoint.

Mohammad Riazi-Kermani
  • 68,728
  • I have troubles showing that the open balls are actually disjoint. Could you elaborate a bit more on this? MeesdeVries already hinted that cutting the radius in half definitely does the trick, but I haven't been able to prove this... – Student Jul 18 '18 at 11:14
  • Found it, a drawing does wonders! – Student Jul 18 '18 at 11:22
  • 1
    In general the open intervals need not be disjoint. Take $E = { 0, 1 }$ and $N_{r_0} = (-1,1), N_{r_1} = (0,2)$. – Paul Frost Jul 18 '18 at 11:28
  • @PaulFrost The open intervals do need to be disjoint but you may pick them disjoint to make sure that the function is one-to -one. – Mohammad Riazi-Kermani Jul 18 '18 at 11:36
  • Does "pick them disjoint" mean that you can somehow construct the $N_{r_x}(x)$ pairwise disjoint? How would that work? – Paul Frost Jul 18 '18 at 12:11
  • The open balls are not disjoint. The intersection of any two of those balls and E is empty which suffices for the injection. – William Elliot Jul 18 '18 at 19:11
  • @WilliamElliot Perhaps I have a mental block, but I still do not see how that could help. You choose rational $q_x \in N_{r_x}(x), q_y \in N_{r_y}(y)$ and have to show that $q_x \ne q_y$ if $x \ne y$. You certainly know that $N_{r_x}(x) \cap N_{r_y}(y) \cap E = \emptyset$, but what does this mean for $q_x,q_y$? It may happen that $q= q_x = q_y \in N_{r_x}(x) \cap N_{r_y}(y)$, but $q\notin E$. – Paul Frost Jul 18 '18 at 21:56
  • See my answer where I show injectivity without using disjoint sets. I dare you to prove the open sets are disjoint. Otherwise your proof fails – William Elliot Jul 20 '18 at 08:14
1

We can take $N_{r_x}(x) = (a_x,b_x)$ with $a_x, b_x \in \mathbb{Q}$. The set $M = \{ (a_x, b_x) \mid x \in E \}$ is countable because there are only countably many intervals with rational endpoints. Write $M = \{ N_i \}$ with finitely many or countably infinitely many $N_i$. We have $N_i \cap E = \{ x_i \}$ with $x_i \in E$. Then $E = \bigcup_{x \in E} N_{r_x}(x) \cap E = \bigcup_i N_i \cap E = \bigcup_i \{ x_i \}$.

Added:

You can also use this to produce an injection $i : E \to \mathbb{Q} \times \mathbb{Q}$. For any $x$ choose $N_{r_x}(x) = (a_x,b_x)$ with $a_x, b_x \in \mathbb{Q}$. Define $i(x) = \langle a_x,b_x \rangle \in \mathbb{Q} \times \mathbb{Q}$ (we use $\langle a_x,b_x \rangle$ to distinguish the pair from the interval $(a_x, b_x)$). If $\langle a_x,b_x \rangle = \langle a_y,b_y \rangle$, then $\{ x \} = (a_x,b_x) \cap E = (a_y,b_y) \cap E =\{ y \}$.

By the way, the above proof is based on the assumption that $E$ has no limit points in $E$ itself. It remains open whether this is the correct interpretation of the question whose prerequisite could also be understood in the sense that $E$ has no limit points in $\mathbb{R}$. What we have proved is certainly the more general case.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125