# Prove that [a, b] of R is closed in topology

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.

• 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 $$\forall x\in U,\ \exists \varepsilon>0,\ \forall y\in (x-\varepsilon,x+\varepsilon),\ y\in U$$
– user239203
Nov 16 '19 at 8:15
• $U$ is open on the real line it $\forall x \in U$, there is some $\delta >0$ such that $(x-\delta, x+\delta) \subset U$. Nov 16 '19 at 8:18
• The usual topology for R is the topology generated by the open intervals. 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.).