Help! I can't think why it has to be closed as well.

63 Views Asked by At

In section 9 of Terence Tao's book, we have the Heine Borel Theorem:-

If $X \subset \mathbb R$, it is bounded and closed iff any sequence of elements in $X$ has a convergent subsequence with limit in $X$

Everything's fine, it has a nice similarity with Bolzano - Weirstraß, but i cannot understand why it has to be closed?


My attempt to prove (the tougher implication) is like this-

Let $X \subset [-M,M]$.Then the sequence $(x_n)_{n=0}^{\infty}$ containing each element of $X$ is bounded, and letting $L^+$ denote the $\limsup (a_n)_{n=0}^{\infty}$ we have $|L^+|<M$, implying the limit superior is finite and thus it has a convergent subsequence with limit $L^+$.

It seems fine to me, but i can't understand why the condition "$X$ is closed" is necessary.

Thanks for helping :)

1

There are 1 best solutions below

10
On BEST ANSWER

Yes, it has a convergent subsequence. And the limit of that subsequence belong to $X$, since $X$ is closed. Otherwise, you cannot assure that that will happen.