# Does $C[0, 1]$ have uncountable disjoint open sets?

Let $$(X, d)$$ be a metric space. Let $$J$$ be an indexing set. Consider a set of the form $$S = \{x_j\in X\mid j\in J\}$$ with the property that $$d(x_j, x_k) = 1$$for all $$j\neq k$$, $$j, k \in J$$.
Which of the following statements are true?
(a) If such a set exists in $$X$$, then there exist open sets $$\{U_j\}$$ in $$X$$ such that $$U_j \cap U_k = \emptyset$$, for all $$j \neq k$$.
(b) There exists such a set $$S$$ in $$C[0, 1]$$ with $$J$$ being uncountable.

My attempt:

Option a is correct. Consider the sets $$U_j= B_\frac{1}{3}(x_j)$$ which are open (singletons are open sets in discrete metric). Clearly $$U_j \cap U_k=\emptyset$$ as $$d(x_j, x_k)=1$$.

How to disprove option b? Is the explanation correct for option a?

• @Saucy O'Path Sorry don't get it.
– PAMG
Commented Mar 14, 2019 at 6:59
• No separable metric space can have uncounatbly many points at distance $1$ from each other. Commented Mar 14, 2019 at 8:00
• I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_{1/3}(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j$ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $\triangle$ inequality, that $B_{1/3}(x_j)$ and $B_{1/3}(x_k)$ are disjoint when $j\ne k$ . Commented Mar 14, 2019 at 9:47

$$C [0,1]$$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $$\{f_n\}$$ be dense and for each $$j \in J$$ pick $$k_j$$ such that $$d(f_{k_j}, x_j) <\frac 1 2$$. Now verify that $$j\in J \to k_j$$ is a one to one map from $$j$$ into the set of positive integers. Hence $$J$$ is necessarily countable.