2
$\begingroup$

Does there exist an uncountable subset $A \subseteq\mathbb R$ such that there exist an injective function $f:A \to \mathbb R$ such that for every $a \in A$ , there exist a connected subset $S \subseteq \mathbb R$ with more than one point such that $\{a\}=f^{-1}(S)$ ? Can $f$ be continuous ?

$\endgroup$
8
  • 1
    $\begingroup$ Your question puts no condition on $S$. Do you mean $\{a\}=f^{-1}(S)$ perhaps? $\endgroup$
    – Lee Mosher
    Jan 18 '16 at 15:34
  • $\begingroup$ @LeeMosher : Yes , it is $S$ ; corrected $\endgroup$
    – user228169
    Jan 18 '16 at 15:37
  • $\begingroup$ @bburGsamohT: by definition, $f^{-1}(S) = \{x \in A \bigm| f(x) \in S\}$. $\endgroup$
    – Lee Mosher
    Jan 18 '16 at 15:42
  • $\begingroup$ @user228169 Ah shoot I always forget about when you need to invoke surjectivity and all that. Sorry. $\endgroup$
    – TomGrubb
    Jan 18 '16 at 15:44
  • 2
    $\begingroup$ @skyking: $f^{-1}(S)$ can't have lower cardinality than $S \cap f(A)$, but it certainly can have lower cardinality than $S$. $\endgroup$
    – Lee Mosher
    Jan 18 '16 at 18:21
2
$\begingroup$

No such setup exists. Here is an argument by contradiction.

We may assume $A \subset \mathbb{R}$ and $f$ is inclusion.

Assume by contradiction that for each $a \in A$ there is an interval $S_a \subset \mathbb{R}$ such that $S_a \cap A = \{a\}$. We may assume, by truncating $S_a$ if necessary, that $a$ is an endpoint of $S_a$. Choose $S'_a$ to be a subinterval of $S_a$ with endpoint $a$ such that $$0 < \text{Length}(S'_a) \le \frac{1}{2} \text{Length}(S_a) $$ It follows that if $a \ne b \in A$ then $S'_a \cap S'_b = \emptyset$. Choose a rational number $r_a \in S'_a$. Then the function $a \mapsto r_a$ is an injective function $A \mapsto \mathbb{Q}$, and therefore $A$ is countable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy