Basic point set topology. Im starting a intro to topology class and there are two separate problems in our notes.

In $X = \Bbb R$,
Consider $T$ to be the collection of all subsets
of the form $(−\infty, a] = \{t ∈ \mathbb{R} \space| \space t ≤ a\}$
Where $a$ is a point in $\Bbb R$ along with $\Bbb R$ and the empty set.
Determine whether
or not $T$ is a topology on $\Bbb{R}$.

It seems that the answer is no.
However in $X = \Bbb R$,
if we consider $T$ to be the collection of all subsets
of the form  $(−\infty, a) = \{t ∈ \mathbb{R} \space| \space t < a\}$ it seems that $T$ is an topology but I don't really understand why or how to show this rigorously.
 A: Let $U_a = (-\infty, a]$ and let $A = (-\infty,0)$.  Then $$\bigcup_{a \in A} U_a = (-\infty, 0)$$ is a union that is not in your topology $T$, giving you a counterexample.
A: Let $T=\{(-\infty,a]\mid a\in\Bbb{R}\}$ then $\bigcup_{n=1}^{\infty}\Big(-\infty,a-\frac1n\Big]=(-\infty,a)\not\in T$, implying $T$ is not a topology.
A: Your second $T$ is a topology: the other answers are wrong or didn't show what OP wanted.


*

*$\varnothing,\mathbb{R}\in T$.

*If you have a collection $\{r_i\}_{i\in I}$ with $r_i\in\mathbb{R}$ for all $i\in I$, then $\bigcup_{i\in I}(-\infty,r_i)=(-\infty,r)$ for $r=\sup_{i\in I}r_i$ (can be shown with very simple set theory: is trivial), so $(-\infty,r)\in T$ (note that adding $\varnothing$ is trivial).

*If you have a finite collection $\{r_i\}_{i\in I}$ with $r_i\in\mathbb{R}$ for all $i\in I$ with $I$ finite, then $\bigcap_{i\in I}(-\infty,r_i)=(-\infty,\min_{i\in I}r_i)\in T$. If you include $\varnothing$, then the intersection is $\varnothing\in T$.
All three properties are true for $T$, so $T$ is a topology.
