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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?

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  • $\begingroup$ @Saucy O'Path Sorry don't get it. $\endgroup$
    – PAMG
    Commented Mar 14, 2019 at 6:59
  • $\begingroup$ No separable metric space can have uncounatbly many points at distance $1$ from each other. $\endgroup$ Commented Mar 14, 2019 at 8:00
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    $\begingroup$ 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 $j\ne k$ . $\endgroup$ Commented Mar 14, 2019 at 9:47

1 Answer 1

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$C [0,1]$ (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.

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