From munkres's topology 17, it shows that the subset [a, b] of $\Bbb R$ is closed because its complement $\Bbb R - [a, b] = (-\infty, a) \bigcup (b, +\infty) $ is open. I don't know why $(-\infty, a) \bigcup (b, +\infty) $ is open because the example of the book does not specify the topology of $\Bbb R$. How can I prove a subset is open or closed or not without knowing the topology? Another example, why subset [a, b) of $\Bbb R$ is neither open nor closed.

They are obvious to me when I use the knowledge I learn in algebra, however, I really could not use the definition of topology to prove it without knowing the topology.

  • $\begingroup$ You cannot prove if a set is closed/open unless you know the topolog, of course. You should read backwards until you find the definition of the stadard topology of $\Bbb R$. A common definition of the standard topology of $\Bbb R$ is that $U$ is open if and only if $$\forall x\in U,\ \exists \varepsilon>0,\ \forall y\in (x-\varepsilon,x+\varepsilon),\ y\in U$$ $\endgroup$
    – user239203
    Nov 16 '19 at 8:15
  • $\begingroup$ $U$ is open on the real line it $\forall x \in U$, there is some $\delta >0$ such that $(x-\delta, x+\delta) \subset U$. $\endgroup$ Nov 16 '19 at 8:18
  • 1
    $\begingroup$ The usual topology for R is the topology generated by the open intervals. $\endgroup$ Nov 16 '19 at 8:41

Munkres discussed the topology of $\Bbb R$ in 14: it's the order topology, on page 85 (2nd edition, top) he even explicitly says under Example 1:

The standard topology on $\Bbb R$, as defined in the preceding section, is just the order topology derived from the usual order on $\Bbb R$.

So he does define it. And $(b , +\infty)$ is a union of open intervals $(b,b+1), (b+\frac12, b+2), (b+\frac32, b+3), \ldots$ (all open intervals are order-topology open) or use the alternative definition that uses all sets of the form $(a,+\infty)$ and $(-\infty,b)$ as a subbase for the order topology (he discusses this on p. 86 in the part under where he defines open rays etc.).


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