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Are family of subsets $\delta$ topology on $R$ number line?

$\delta$={${\emptyset},R,(-\infty,x),x\in R$}

$\delta$={${\emptyset},R,(-\infty,x),x\in Z$}

$\delta$={${\emptyset},R,(-\infty,x),x\in Q$}

In order to be topology it must fulfil $3$ criterion.

$1$)$\emptyset$ and $R$ must $\in$ $\delta$

$2$)Any number union of elements in $\delta$ must $\in$ $\delta$

$3$)Any finite number intersection of elements in $\delta$ must $\in$ $\delta$.

For every of $3$ examples first criterion holds.

For second criterion if we take $(-\infty,x)x\in R$ union will be some $(-\infty,x)$ where still $x\in R$

same for intersection and $Q$ and $Z$.

In my work 3 examples are topology.

Can you explain what is difference here because I think that I am doing something wrong here.

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1 Answer 1

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If $A=\{x\in\Bbb Q\mid x<\sqrt2\}$, then, for each $x\in A$, $(-\infty,a)$ belongs to your third set. However$$\bigcup_{x\in A}(-\infty,x)=\left(-\infty,\sqrt2\right),$$which does not belong to it. Therefore, the third set is not a topology on $\Bbb R$.

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  • $\begingroup$ If @JoséCarlosSantos it is same problem but $(-\infty,x]$ your solution still works? $\endgroup$
    – unit 1991
    Oct 28, 2021 at 17:40
  • $\begingroup$ Yes, it does. Do you think that I've answered your question? $\endgroup$ Oct 28, 2021 at 17:43
  • $\begingroup$ Yes,thank you.. $\endgroup$
    – unit 1991
    Oct 28, 2021 at 17:46
  • $\begingroup$ I'm glad I could help. $\endgroup$ Oct 28, 2021 at 17:47

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