2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What are the definitions of $\mathbb{R_L}$ and $\mathbb{R_k}$? $\endgroup$ – bof Aug 3 '14 at 8:36
  • 1
    $\begingroup$ @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 $b$. $\endgroup$ – Fawzy Hegab Aug 3 '14 at 13:21
4
$\begingroup$

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 $\mathbb R_K$.

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Fawzy Hegab Aug 3 '14 at 13:33
  • 1
    $\begingroup$ @MathsLover thanks for pointing this out, that was indeed confusing and careless notation. $\endgroup$ – vociferous_rutabaga Aug 3 '14 at 13:53
  • 1
    $\begingroup$ Never mind, for me it wasn't confusing as I understood that you mean $K$. It's all about "Future Visitors!". $\endgroup$ – Fawzy Hegab Aug 3 '14 at 15:25
6
$\begingroup$

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 $\mathbb{R}_L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.