Let $\mathcal{A}=\{(a, \infty) \mid a < 0 \} \cup \{(-\infty, b) \mid b > 0\}$. Show that $\mathcal{A}$ is a subbasis for some topology on $\Bbb R$.

Let $\mathcal{A}=\{(a, \infty) \mid a < 0 \} \cup \{(-\infty, b) \mid b > 0\}$. Show that $\mathcal{A}$ is a subbasis for some topology on $\Bbb R$. Find the interior of $[0,1]$ and $[-1,1]$.

$\mathcal{A}$ is a subbasis for some topology $\tau$ if the finite intersections of the elements of $\mathcal{A}$ form a basis for $\tau$.

The only possible elements we can create from intersecting elements of $\mathcal{A}$ are $$(a,b) \\(a,\infty) \\(-\infty, b)$$

so this is a subbasis for the usual topology on $\Bbb R$? Do the sets $(a,\infty)$ and $(-\infty, b)$ belong in the usual topology of $\Bbb R$?

The interior of $[0,1]$ would then be $\emptyset$ as $0$ is not in any of the intervals, but what is the interior of $[-1,1]$? It's the largest open set contained in $[-1,1]$ but I don't know wheter it is $(-1,1)$.

Solution 1:

First, any family of sets with union containing $\mathbb R$ is a subbasis for some topology on $\mathbb R$, so the first part is trivial.

We can't get any arbitrary interval $(a, b)$ as a finite intersection of subbase elements, or as union of intersections, so the topology isn't standard.

Finite intersection of elements of $\mathcal A$ is either empty, or $(a, b)$ where $a < 0$ or $a = -\infty$ and $b > 0$ or $b = \infty$. It's simply to check that any such set can be obtained as intersection of elements from $\mathcal A$, so the topology generated by $\mathcal A$ is "open sets are empty and (possibly infinite) intervals containing $0$".

From this we indeed see that interior of $[0, 1]$ is empty - any non-empty open set contains some negative numbers - and interior of $[-1, 1]$ is $(-1, 1)$ - any open set containing $-1$ contains also some smaller numbers, and any open set containing $1$ contains also some larger.