Let $V$ be a normed linear space and $D$ be a dense subset of $V.$ Let $x \in D.$ My question is $:$
Is $D \setminus \{x\}$ dense in $V\ $?
Clearly, $V \setminus \{x\} \subseteq \overline {D \setminus \{x\}}$ because for any $V \ni y \neq x$ and any open neighbourhood $U_y$ around $y$ there exists $r \gt 0$ such that $B(y; r) \subseteq U.$ Let $s : = \text {min} \left \{r, \|x - y \| \right \}.$ Then $B(y; s) \subseteq B(y; r) \subseteq U_y.$ Now $D$ being dense we have $B (y; s) \cap D \neq \varnothing.$ But $x \notin B(y; s)$ (for otherwise $\|x - y\| \lt s,$ a contradiction to our assumption) and hence $B(y; s) \cap (D \setminus \{x \}) \neq \varnothing$ and hence $U_y \cap \left (D \setminus \{x\} \right ) \neq \varnothing,$ which is what our claim was.
and, since $V$ doesn't have any a isolated point of $V$ we are through so $x$ is a limit point of $D \setminus \{x\}$ as well.