Standard topology on $\mathbb{R}$ is not finer than $K-$ topology on $\mathbb{R}$.

Question

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 $B \not = (-2,1)$

Answer 1

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.

Answer 2

The book might have been more precise on the definition of the K-topology.

K-topology is generated by basis elements of two types. The first is $(a,b)$ that is the same basis elements of the Standard topology.

The second type is $(a,b) - K $.

The way to check is to pick one basis element of $\mathcal{R}_K$ and one point in it. Then try to find one basis element in the second topology that will include the same point and fit into the same basis element of $\mathcal{R}_K$.

We try

$$0 \in (-1,1)-K$$

and see there is no interval in the standard topology of the form $(x,y)$ that would include zero and fit inside $(-1,1)-K$. Because it will inevitably grab some point of K that lie in $(0,1)$.

Your construction does not work because you try to construct a 'cover' by taking more than one basis element from the standard topology $\mathcal{R}$ and taking their union.

You need to be able to pick exactly ONE basis element.