# Are $\mathbb{R_k}$ and $\mathbb{R_L}$ really non-comparable topologies?

First of all, we notice that: $$(a,b)=\bigcup_{a<x<b} [x,b).$$

Also, we notice that: $$(0,1)-\left\{\frac{1}{n} |\, n\in \mathbb{Z^+}\right\}=\left(\frac{1}{2},1\right)\cup \left(\frac{1}{3},\frac{1}{2}\right)\cup \left(\frac{1}{4},\frac{1}{3}\right)\cup \left(\frac{1}{5},\frac{1}{4}\right)\cup \ldots =$$

$$\bigcup_{a_i=1}^{\infty} \left\{ \left(\frac{1}{2a_i},\frac{1}{2a_i-1}\right)\bigcup\left(\frac{1}{2a_i+1},\frac{1}{2a_i}\right)\right\}$$

We also can express every basis element of $\mathbb{R_k}$ of the form $(a,b)-K$ in a similar way to $(0,1)$.

So every element of $\mathbb{R_k}$ can be expressed as a union of open intervals. But, we can express any open interval as a union of intervals of the form $[x,b)$ as we have done in the beginning . So, we conclude that every element in $\mathbb{R_k}$ is expressible using elements of $\mathbb{R_L}$. So $\mathbb{R_L}$ is finer than $\mathbb{R_k}$.

My question is, what is the wrong with my presentation? I know that there is a proof for that both topologies are not comparable and it convinced me but I can't find where my way of thinking went wrong, any help is appreciated.

• What are the definitions of $\mathbb{R_L}$ and $\mathbb{R_k}$? – bof Aug 3 '14 at 8:36
• @bof , $\mathbb{R_L}$ is the topology whose basis elements are all intervals are of the form $[a,b)$ for $a$ and $b$ are real numbers. Let $k=\text{{$\frac{1}{n}|n\in \mathbb{Z^+}$}}$, the topology $\mathbb{R_k}$ is the topology whose basis elements are the union of $1$) all the intervals of the from $(a,b)$ for real numbers $a$ and $b$, and $2$) all the sets of the form $(a,b)-K$ for real numbers $a$ and $b$. – Fawzy Hegab Aug 3 '14 at 13:21

The issue lies in the assertion that any interval $(a,b) - K$ can be decomposed as a union of open intervals "in a way similar to $(0,1)$." Indeed, the procedure does not extend if $a<0<b$. Let's assume $b=1$, and $a<0$. Note that $(a,b)-K$ is not equal to $(a,0)\cup\big(\cup_{i=1}^\infty (\frac{1}{i+1},\frac{1}{i})\big)$, since no interval of the form $(\frac{1}{i+1},\frac{1}{i})$ contains zero.

However, your investigation shows that the lower limit topology on $\mathbb R$ is "almost" finer than the $K$-topology: all open sets of $\mathbb R_K$ not containing $0$ are also open in $\mathbb R_L$. This reflects the fact that the induced K-topology on $\mathbb R_K -\{0\}$ coincides with the usual subspace topology on $\mathbb R-\{0\}$.

For posterity, here are a few words about how to show that the two topologies are incomparable (although I understand that you are already convinced). First check that $K=\{\frac{1}{n} : n \in \mathbb N \}$ is closed in $\mathbb R_K$ but not in $\mathbb R_L$. To finish up, observe that $[0,1)$ is not open in $\mathbb R_K$.

• Thank you very much :) but I think it's better to use $K$ instead of $N$ so, the future reader of the question doesn't confuse what is that $N$?.@Morgan O – Fawzy Hegab Aug 3 '14 at 13:33
• @MathsLover thanks for pointing this out, that was indeed confusing and careless notation. – vociferous_rutabaga Aug 3 '14 at 13:53
• Never mind, for me it wasn't confusing as I understood that you mean $K$. It's all about "Future Visitors!". – Fawzy Hegab Aug 3 '14 at 15:25

It's true that $(0,1) \setminus K$ can be written as a union of open intervals, but it's not true that you can do it for all sets of the form $(a,b) \setminus K$. For example $$(-1, 1) \setminus K = (-1, 0] \cup \bigcup_{n\geq 1} (\frac{1}{n+1}, \frac{1}{n})$$

Note that $(-1, 0]$ is not open in $\mathbb{R}_L$.