Discussion between Henno Brandsma and unit 1991

4:45 PM

-1

Q: $\delta=$ {$\emptyset,R,{(-x;x)|x\in R,x\gt0}$} topology on real number line.

$\delta=$ {$\emptyset,R,{(-x;x)|x\in R,x\gt0}$} $\delta=$ {$\emptyset,R,{(-x;x)|x\in Q,x\gt0}$} $\delta=$ {$\emptyset,R,{(-x;x)|x\in Z,x\gt0}$} Are family $\delta$ of subsets topology on real number line. For three examples $R$ and $\emptyset$ are there so first axiom holds. If we take union of $...

unit 1991

Similar question does not answer my question.

Henno Brandsma

Your union arguments make no sense. The other question shows 2, the rational endpoints, is not a topology. The union of a subfamily is not always $\Bbb R$.

unit 1991

@HennoBrandsma That's what I was asking.I thought that my work is not correct so I posted here.

Henno Brandsma

If $I \subseteq \Bbb R^+$ the question is what is $\bigcup\{(-x,x)\mid x \in I\}$ for the first candidate, and the union axiom.

unit 1991

@HennoBrandsma For $Z$ the answer of similar queston does not work.

Henno Brandsma

4:45 PM

The question is what is the difference between these cases?

unit 1991

@HennoBrandsma Yes.

Henno Brandsma

Can you answer the question I asked about case 1? With $I$?

unit 1991

For first candidate when $x \in R$ union is R?

If I understood correctly for $x \in Z$ union is also $R$ but for $\Q$ there is an example which we get one endpoint irrational.

Henno Brandsma

Show that for the $\Bbb Q$ and $\Bbb R$ case we have $\bigcup_{x \in I} (-x,x)=(-\sup I, \sup I)$ which shows how it fails for $\Bbb Q$, for $\Bbb Z$ we have the same identity and there $\sup I=+\infty$ or $\max I$.

unit 1991

4:58 PM

Is there an easy way to show that $\bigcup_{x \in I} (-x,x)=(-\sup I, \sup I)$.So we need to show that each set is subset of another. We take $x_1$ from $\bigcup_{x \in I} (-x,x)$ let's say $I$ is $Q$ but $Q$ has a supremum?

Henno Brandsma

5:10 PM

The supremum of the whole set $\Bbb Q$ would be $+\infty$, but it can be be any real number $>0$ for suitably chosen $I$.

You just show two inclusions

In all cases, if $\sup I=+\infty$ the union is $\Bbb R \in \tau$.

But it need not be.

unit 1991

Really thanks for help I'll try to think more about what you said.

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