In our real analysis class we are working through 'Real and Complex Analysis' by Rudin and covered topological spaces (but not bases, subbases and other ways of generating topologies, so I can't use these in the exercise).
Let $\tau =$ the collection of sets $(a,b),[-\infty,a),(a,\infty]$ and any union of these types. Show that $\tau$ is a topology.
Here's my approach:
- $\mathbb{R} \in \tau$ since $\mathbb{R} = [-\infty,a)\cup(a,\infty]$. $\emptyset \in \tau$.
- For any $A_1,A_2,\ldots \in \tau,$ we have $\bigcup_{i \in I} A_i = \bigcup_{i \in I} (a_i,b_i) \in \tau.$
- For any $A_1,\ldots,A_n \in \tau$, we have $\bigcap_{i=1}^n (a_i,b_i) \in \tau.$
However, I feel like my approach is too naive and I'm missing some details.
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$\begingroup$Ok here is my attempt at another answer, please comment if there are still some parts missing/wrong.
Let $\overline{\mathbb{R}} = [-\infty,\infty]$ and let $b <a$. Then we have $\overline{\mathbb{R}} = [-\infty,a) \cup (b,\infty] \in \tau$. Next, let $a =b$ so that $(a,b) = \emptyset \in \tau$.
Now, we define $A = \bigcup_{j \in J} A_j = \bigcup_{j \in J}\left(\bigcup_{i \in I}(a_i,b_i)\right)$. Then $A \in \tau$ since the unions of the unions of intervals is a union of intervals, which are in $\tau$.
Last, we define $A = \bigcap_{j \in J} A_j = \bigcap_{j \in J} \left(\bigcup_{i \in I}(a_i,b_i)\right).$ Then, for any $x \in \bigcup_{i \in I} (a_i,b_i),$ $x \in A$ and so A is open. (this doesn't feel correct..)
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