Show that the standard topology on $\mathbb{R}$ is not finer than $K-$ topology on $\mathbb{R}$.
An equivalent statement is to show that $\exists \,x\in \mathbb{R} $ and $\exists$ a basis element $B$ in $K-$ topology such that for all open sets $(a,b)$ in standard topology containing $x$, $(a,b) \not\subset B.$
Munkres suggests considering the point $0 \in \mathbb{R}$ and $B = (-1,1)-K$ and claims that no open interval $(a,b)$ which contains $0$ is a subset of $B.$
I have trouble following the claim. I am a bit unsure as to how the set $B = (-1,1)-K$ looks. Because if $K = \{\frac{1}{n} : n\in \mathbb{N}\}$ then
$(-1,1)-1=(-2,0)$
$(-1,1)-1/2=(-1.5,0.5)$
$(-1,1)-1/3=(-2/3,2/3)$
$\vdots$
$(-1,1)-1/n $ approaches $(-1,1)$ when $n$ is very large.
Sketching each of these intervals looks to me the set $B$ is nothing but the set $(-2,1).$ But that is not the case since we would have $0\in(-1/2,1/2)$ and $(-1/2,1/2)\subset B.$
Can anyone point out why $B \not = (-2,1)$
Best Answer
The minus symbol in the expression $(a,b)-K$ denotes relative complement, not element-wise subtraction.
Thus \begin{align*} (-1,1)-K &=(-1,1)\setminus K\\[4pt] &= (-1,0] \cup\left( \bigl({\small{\frac{1}{2}}},1\bigr) \cup \bigl({\small{\frac{1}{3}}},{\small{\frac{1}{2}}}\bigr) \cup \bigl({\small{\frac{1}{4}}},{\small{\frac{1}{3}}}\bigr) \cup \cdots \right) \end{align*} By definition, any open interval is open in the $K$-topology, hence any open set in the standard topology is still open in the $K$-topology.
To see that the $K$-topology is strictly finer than the standard topology, simply note that no open interval containing the element $0$ is a subset of $(1,1)-K$, so $(1,1)-K$ is not open in the standard topology.