Existence of a special kind of continuous injective function $f\colon A \to \mathbb R$, where $A$ is countable, relating to connectedness Let $A \subseteq \mathbb R$ be a countable set ($A$ induced with usual subspace topology), then does there necessarily exist a continuous injective function $f\colon A \to \mathbb R$ such that for every $a \in A$, there exist a connected subset $S\subseteq \mathbb R$ (with more than one point) such that $\{a\}=f^{-1}(S)$ ?I can prove the existence of such function if continuity is not required; if continuity is also required I am totally stuck. Please help. Thanks in advance

Look at $ℚ \subset ℝ$, there is no continuous injective map $f:ℚ →ℝ$ which fulfills your conditions.
Lemma: If $f^{-1}(S) = \{a \in ℚ\}$ and $f(a), f(a)+ε \in S$ for $ε > 0$ and $S$ connected then $f(A) \subset (-∞, f(a)]$.
Proof: As both $(-∞, a)\cap ℚ$ and $(a, ∞)\cap ℚ$ are connected, their images must each lay in one connected component of $ℚ$. Some number larger than $a$ and some number smaller than $a$ must have it's image under f closer than $ε$ to a, thus in $(-∞, f(a))$. Consequently, $f(A) \subset (-∞, f(a)]$.
A similar statement holds for $f(a'), f(a')-ε \in S$, then $f(A) \subset [f(a'), ∞)$.
In conclusion, any continuous map $f:ℚ → ℝ$ has at most two values $q, q' \in ℚ$ for which your condition holds.
