prove that the following collection of subsets of $\Bbb R$ is topology

Question

prove that the following collection of subsets of $R$ is topology: topology consists of $R$, $\emptyset$ and every interval $[n,\infty]$, where $n$ is a positive integer.

This is the exercise from the e-book "Topology without tears". I think this exercise is wrong since there is a union of these intervals which is equal $(\inf(a+$1/n$),\infty)$. The $\inf$ does not belong to the set of intervals like [-n,n].

I'm I wrong?

Answer 1

The point is that $n$ can only be a positive integer, according to the exercise.

Intersections of two such sets are of the same form: $[n_1,\infty) \cap [n_2, \infty) = [\max(n_1,n_2), \infty)$ and by induction the collection is then closed under finite intersections too.

If $O_i, i \in I$ is a collection of open sets, if some $O_i= \mathbb{R}$ the union is $\mathbb{R} \in \mathcal{T}$, so we assume no $O_i = \mathbb{R}$. If $O_i = \emptyset$ we can omit it, as it has no effect on the union. So we can assume all $O_i = [n_i, \infty)$ where $n_i \in \mathbb{N}^+$. Any non-empty subset of the natural numbers has a minimum, so say $n_{i_0} = \min \{n_i: i \in I\}$ and note that $\cup_i O_i = [n_{i_0}, \infty) \in \mathcal{T}$. So $\mathcal{T}$ is closed under all unions.

Answer 2

The exercise is correct. Pick any collection of integers $n$. If it has a smallest element $n_0$, the union of all the $[n,\infty)$ is just $[n_0, \infty)$, otherwise it is $(-\infty,\infty) = \mathbb{R}$. Analogously, the intersection of finitely many subsets $[n_i,\infty)$ is just $[N,\infty)$, where $N:=\max\{n_i\}$.

Answer 3

I'm assuming $n\in\mathbb{Z} $. If so, then you just need to check the definition of Topology. Let $\mathcal{T} $ be the set you described.

$\mathbb{R} \in \mathcal{T}, \emptyset \in \mathcal{T} $. Take two intervals $I_a=[a, \infty[$ and $I_b=[b, \infty[$, $a, b\in\mathbb{Z} $ and $a\le b$. Their union is $I_a$, their intersection is $I_b$. Infinite unions of these intervals yield either another interval of the same kind, except the union of all intervals $I_c$ where $c$ is a negative number. This yields: $$\bigcup_c I_c =\quad ]-\infty, \infty[ \quad= \mathbb{R} $$ Then, any union of intervals belong to $\mathcal{T}$, finite intersections of intervals belong to $\mathcal{T}$.

Thus, by the definition of topology, $\mathcal{T} $ is a topology for $\mathbb{R} $.

EDIT: As noted in the comments, $n\in\mathbb{N} $. However, this is a particular case of the above, where any interval has a positive nonzero lower bound. That also means: $$\bigcup_c I_c =\quad [1, \infty[ $$ Where $c\in\mathbb{N} $. (depending on whether you like putting 1 or 0 as the first number of the set of the naturals or not)

The resulting collection of sets is again a suitable topology for $\mathbb{R}$.