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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.

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  • $\begingroup$ Maybe try to relate II and III. Why do you think that "the rest are true"? $\endgroup$ – Mirko Oct 15 '19 at 4:34
  • $\begingroup$ "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. $\endgroup$ – Henno Brandsma Oct 15 '19 at 4:44
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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.)

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.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 -$\infty$.

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