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?

  • $\begingroup$ 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

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


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


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