Let $A$ and $B$ be two non-empty algebraic subsets of $\mathbb{C}^n$ such that $B$ is strictly contained in $A$. I am trying to show that the difference $A-B$, which is a semi-algebraic set, is not closed for the Euclidean topology. Of course if $A$ is finite then the answer is negative. However, if we suppose that the interior of $A$ is non-empty, can we show that it is not closed?
Thanks in advance!
On
If $A-B$ denotes set difference $A \setminus B$, let $A$,$B$ be closed and suppose $A\setminus B$ is closed: this happens precisely when $B$ is open in $A$. So there are plenty of positive examples: if $A$ is closed and discrete, for any subset $B$ of $A$ we have that $B$ is closed and $A \setminus B$ is closed (e.g. $A = \mathbb{Z} \subseteq \mathbb{R}$ would work. But for $A =X$ we get that if $A \setminus B$ is closed and $B$ is closed, $B$ would be clopen and so in Euclidean spaces (or any connected $X$) this would happen iff $B =\emptyset$ or $B=X$. So it could be closed or not, the most "likely" being not closed.
Consider $\mathbb{C}$, The real numbers $\mathbb{R}$ is a closed subset of $\mathbb{C}$, and the single point $x=(0,0)$ is a closed subset of the $\mathbb{C}$ contained in $\mathbb{R}$, remove it from $\mathbb{R}$, then the remaining set is not closed as $x$ would be a limit point of it, not in that set!