Will the dense subset of an uncountable set remain dense if we remove a single element? [closed]

Question

Let $A$ a dense subset of an uncountable topological space $X$ .
Would $A$ remain dense if we removed a single element from it?
Is the case same for both Hausdorff and non-Hausdorff spaces?

Answer 1

Consider an uncountable set $X$ with discrete topology. $X$ is a dense subset of itself. Note $X \setminus \{x\}$ (remove $x$ from $X$) is closed in $X$ and so not dense in $X$.

Answer 2

In the Zariski topology on $\mathrm{Spec}(\mathbb{R}[X])$ (an uncountable set) the set $A=\{(0)\}$ consisting of a single point is dense.

Answer 3

Given an element $x$, the set of sets that contain $x$, combined with the empty set, forms a topology. That is, $\tau = \{S: (x \in S) \lor (S= \emptyset\}$ $^{[1]}$. Then $\{x\}$ is a dense subset of $X$, but $\{x\}\setminus \{x\}=\emptyset$, which is not dense.

[1] We could also represent is as $\{S: (y\in S) \rightarrow (x\in S)\}$.