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

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?

• The phrase "dense subset of a set $X$" is ambiguous - the meaning of "dense" depends on the kind of structure $X$ has. If $X$ were a totally ordered set, a "dense subset" would mean something different (although related). I strongly recommend changing the first sentence to "Let $A$ be a dense subset of an uncountable topological space $X$". Commented Aug 4 at 17:26

## 3 Answers

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 $$X$$.

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

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 $$\{S: (y\in S) \rightarrow (x\in S)\}$$.