In K-topology, Find $K'$.

Question

In K-topology, Find $K'$.

Let $x \in \mathbb{R}$.

• If $x<0$, there exists a neighborhood of $x$, which is $(x-1,0)$, such that $(x-1,0)\cap K=\emptyset$. Thus $x\notin K’$ for all $x<0$.

• If $x=0$, there exists a neighborhood of $0$, which is $V=(-1,1)\setminus K$, such that $V\cap K=\emptyset$. Thus $0\notin K’$.

• If $0<x<1$, there exists a neighborhood of $x$, $U=(0,1)\setminus K$, such that $U\cap (K\setminus \lbrace x \rbrace)=\emptyset$. Thus $x\notin K’$.

• If $x=1$, we have that $(\frac{3}{4},2)$ is a neighborhood of $1$, and $(\frac{3}{4},2)\cap (K\setminus \lbrace 1 \rbrace )=\emptyset$. Hence $1 \notin K’$.

• If $x>1$, there exists a neighborhood of $x$, which is $(\frac{x+1}{2},x+1)$, and $(\frac{x+1}{2},x+1) \cap K=\emptyset$. Then $x\notin K'$.

Is that true, please?

Answer 1

In the $K$-topology $K$ is by definition closed, so $K' \subset K$. But it's also clear that $K'=\{0\}$ in the usual topology, which is coarser, and so can only have more limit points than the $K$-topology, and $0 \notin K$. So $K'= \emptyset$ in the $K$-topology.