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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:

  1. $\mathbb{R} \in \tau$ since $\mathbb{R} = [-\infty,a)\cup(a,\infty]$. $\emptyset \in \tau$.
  2. For any $A_1,A_2,\ldots \in \tau,$ we have $\bigcup_{i \in I} A_i = \bigcup_{i \in I} (a_i,b_i) \in \tau.$
  3. 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.

Olorun
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  • The extended real line is $X = \mathbb{R} \cup {\pm \infty}$? If so, then you have to replace $\mathbb{R}$ in 1. with $X$ and this would almost work (you are missing $a$ on the left hand side in 1. but this is easily fixed by taking $[-\infty,1) \cup (0,\infty]$ or something similar). In 2. and 3. you took $A_i$ to be some interval $(a_i,b_i)$ which you cannot assume as $\tau$ contains many sets which are not intervals but union of intervals. The union of the union of intervals is still a union of intervals, so 2. is still easy. But in 3. you really have to prove something which you did not. – Matthias Klupsch Mar 09 '15 at 08:12
  • This is strange, as $(-\infty,\infty)$ is just $\mathbb{R}$. Are you sure you do not mean $[-\infty,\infty]$ in the sense of $\mathbb{R} \cup {-\infty,\infty}$? – Matthias Klupsch Mar 09 '15 at 08:21
  • I might have made a mistake when taking notes since I was sitting in the back and from afar the brackets looks similar. In the book, $[-\infty,\infty]$ is used to that appears to be the case. – Olorun Mar 09 '15 at 08:24
  • Could you please elaborate on this part: "But in 3. you really have to prove something which you did not"? – Olorun Mar 09 '15 at 12:09
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    The problem in 3. is to show that the intersection of (unions of) intervals at hand is contained in $\tau$, that is, to prove that this intersection of (unions of) intervals is a union of intervals. – Matthias Klupsch Mar 09 '15 at 12:26
  • I would be very grateful if you could take a look at my answer. – Olorun Mar 10 '15 at 11:07
  • In general, this type of construction is called the order topology. You can define it on any totally ordered set. – Alan Yan Sep 05 '20 at 23:40

1 Answers1

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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..)

Olorun
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    Your $A_j$ are still not arbitrary elements of $\tau$. Note that a typical element of $\tau$ is of the form $\bigcup_{r} [-\infty,x_r) \cup \bigcup_{s} (a_s,b_s) \cup \bigcup_{t} (y_t , \infty]$, so you have to take your $A_j$ to be of this form. Also note that the index sets used to describe $A_j$ as a union of intervals depends on $j$. As for the part concerning the finite (!) intersection of the $A_j$, you are supposed to show that for any $x \in \bigcap_{j} A_j$ there exists an interval $[-\infty,a_x)$, $(b_x,c_x)$ or $(d_x,\infty]$ containing $x$ and being contained in $\bigcap_{j} A_j$. – Matthias Klupsch Mar 10 '15 at 11:30