# Properties of the topology generated by $\Gamma=\{(a,\infty):a\in \mathbb {R}\}$

Endow $$\mathbb {R}$$ with the right topology generated by $$\Gamma=\{(a,\infty):a\in \mathbb {R}\}$$, and call this space $$X$$. Which of the following is false?

I. $$X$$ is $$\sigma$$-compact (it is the union of countably many compact subsets)

II. $$X$$ is sequentially compact (every sequence has a convergent subsequence)

III. $$X$$ is limit point compact (every infinite subset has a limit point in $$X$$)

IV. $$X$$ is Lindelöf (every open cover of $$X$$ has a countable subcover)

I know that II is false (for example, $$\{-n\}_{n\in \mathbb {N}}$$ does not converge), but I don't know why the rest are true. Thank you.

• Maybe try to relate II and III. Why do you think that "the rest are true"? – Mirko Oct 15 '19 at 4:34
• "doesn't converge" is not enough to refute II, you need to see there is no convergent subsequence as well, though the argument for that is almost identical. – Henno Brandsma Oct 15 '19 at 4:44

A subspace $$A$$ is compact (in this topology) if(f) it has a minimum $$m$$: an element of an open cover containing $$m$$ is of the form $$(m',+\infty)$$ with $$m’ < m \le A$$ and then contains all of $$A$$ too. (the other direction is also true but not needed)

So the simple fact that $$\Bbb R = \bigcup\{ [n, +\infty): n \in \Bbb Z\}$$ shows that $$\Bbb R$$ is $$\sigma$$-compact.

Always $$\sigma$$-compact implies Lindelöf. (countably many finite subcovers give a countable subcover) So I and IV hold.

The sequence $$x_n = -n$$ shows that II does not hold. Any point $$a$$ has a neighbourhood $$(a-1,+\infty)$$ that only contains finitely elements of the sequence so is not a subsequential limit of it.

And if $$A$$ is any infinite set and it has a lower bound $$L$$ then any neighbourhood of $$L$$ contains all of $$A$$, so $$L$$ is a limit point of $$A$$. Otherwise $$A$$ is unbounded below, and for $$a \in A$$ we can find $$b \in \Bbb R$$ with $$b < a$$ and any neighbourhood of $$b$$ intersects $$A$$ in $$a\neq b$$. Such a $$b$$ thus also is a limit point of $$A$$. So $$\Bbb R$$ is limit point compact in this topology.

So only II is false (the formulation suggests that this was a MC question, so you could have stopped at noting II fails for your sequence, but it's good to be curious.)

.1. Show for all a, [a,$$\infty$$) is compact.
.4. Show that if { ($$a_j$$, $$\infty$$) : j in J } is a cover,
then there is a sequence within J that diverges to -$$\infty$$.