Which of the following statements concerning metric spaces are true?

Question

Let $(X,d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = \{x_j \in X\ |\ j \in J \}$ with the property that $d(x_j,x_k) = 1\ $ for all $j \neq k,\ j,k \in J.$ Which of the following staements are true?

a. If such a set exists in $X,$ then there exist open sets $\{U_j \}_{j \in J}$ in $X$ such that $U_j \cap U_k = \emptyset\ $ for all $j \neq k,\ j,k \in J.$

b. There exists such a set $S$ in $\mathcal C [0,1]$ with $J$ being uncountable.

c. If such a set exists in $X,$ and if $X$ is compact, then $J$ must be finite.

Clearly (a) is true. Just taking $0 \lt \varepsilon \leq \frac 1 2$ we can see that the indexed collection of open sets $\left \{B(x_j, \varepsilon) \right \}_{j \in J}$ has the required property mentioned in option (a).

How do I proceed to prove or disprove (b) and (c)? Any help in this regard will be highly appreciated.

Thanks in advance for sparing your valuable time for reading.

Answer 1

b. Assuming that your distance here is $d(f,g)=\sup|f-g|$, then there is no such set. That so because $\mathcal C[0,1]$ is separable; for instance, the set $Q$ of all polynomial functions from $[0,1]$ in $\Bbb R$ with rational coefficients is dense in $\mathcal C[0,1]$. If $J$ existed, for each $j\in J$ there would be an element of $Q$ in $B_{1/2}(j)$. THat's impossible, since those balls don't intersect and there are uncountably many balls.

c. True. If $J$ was infinite, it would have a countable subset. So, we would have a seuence $(x_n)_{n\in\Bbb N}$ of elements of $X$ such that the distance between any two distinct terms would be equal to $1$. Such a sequence has no Cauchy subsequence and therefore no convergent subsequence.

Answer 2

Answer for b): By Wierstsrass Approximation Theorem it follows that polynomials with rational coefficients are dense in $C[0,1]$. This makes the space separable.

Lemma

If $(X,d) $ is separable metric space then we cannot have uncountably many points $(x_j)_{j \in J}$ such that the distance between any two of these is $1$.

Proof: Let $(d_n)$ be a countable dense set in $X$. In $B(x_j,\frac1 2)$ there exists one element of the countable dense set. Call it $d_{n_j}$ Then $j \in J \to d_{n_j}$ is a one-to-one map from $J$ into a countable set so $J$ is necessarily countable.