0
$\begingroup$

I want to show that τ $= \{\emptyset, \mathbb{R}\} \cup \{(-\infty,x]:x\in\mathbb{R}\}$ is not a topology on $\mathbb{R}$, and I assume it has to do with the fact that $(-\infty,x]$ includes $x$.

Clearly $\tau$ contains $\emptyset$ and $\mathbb{R}.$ The intersection of 2 elements of the form $(-\infty,a] \cap (-\infty, b]$ for $a<b\in\mathbb{R}$ is just $(-\infty,a]\in \tau$, so that's okay, as too is any self-intersections and/or intersections involving $\mathbb{R}$ and $\emptyset.$

Any union including $\mathbb{R}$ will give $\mathbb{R}\in\tau$ so assume $\mathbb{R}$ is not included. Any finite union of elements of the form $$(-\infty,a_1]\cup (-\infty,a_2] \cup \dots \cup (-\infty,a_n] \,\text{ for }\, a_1<a_2<\dots<a_n\in\mathbb{R}$$ is just $(-\infty,a_n]\in\tau$ (also if we included $\emptyset$ in the union). And if we had an infinite union then it would again be something of the form $(-\infty,x]\in\tau$ or $(-\infty,\infty)=\mathbb{R}\in\tau$, so again no problem.

What's the issue?

$\endgroup$
  • $\begingroup$ Please explain how you got that last part: "And if we had an infinite union..." $\endgroup$ – bof Jan 16 '18 at 10:53
  • $\begingroup$ @bof I was just thinking of something like $(-\infty,0]\cup\dots\cup (-\infty,1]$ which is an infinite union that I thought gave $(-\infty,1]$, but I guess that's the issue: we never reach $1$ from $0$ so it's $(-\infty,1)$ as the union? Or is that only in the case of a limiting example as in the answers below? $\endgroup$ – Alex.F Jan 16 '18 at 11:17
5
$\begingroup$

The set $\tau$ is not closed for arbitrary unions. For instance:$$\bigcup_{n\in\mathbb N}\left(-\infty,-\frac1n\right]=(-\infty,0).$$

$\endgroup$
3
$\begingroup$

Note that: $$\bigcup_{n=1}^{\infty}\left(-\infty,1-\frac1n\right]=(-\infty,1)\notin\tau$$

$\endgroup$

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.