#15 | Real Analysis | Setting up the Heine-Borel theorem | Hunter College

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[@aninditabanerjee6862]

4
By Bolzano-Weierstrass, any bounded sequence (xk)
in Rn
has a convergent subsequence. Now try to construct a sequence sequence that does not converge, assuming a set is closed and bounded, yet there exists an open covering that has no finite subcover.
Spoiler:
Let A⊂Rn
be closed and bounded, s.t. there exists an open covering which has no finite subcovering. So let Ui
be such an open covering. Pick an open set, which we denote by U0
. Let
x0∈A−U0
, which exists by assumption.
Choose U1∈(Ui)
with x0∈U1
. This exists by assumption.
Now we proceed in that manner and choose a sequence of open sets from the covering and a sequence in A
that is always one step before the covering sequence.
Then the constructed sequence (xk)
is bounded, as it is contained in A
. It has a convergent subsequence, say, to x
. Choose such a sequence and keep the label xk
. As A
is closed, x∈A
.
By assumption, we can choose some V
which covers x
. We do so at any point of the sequence, say after some fixed number of steps. We strip off all elements of the sequence which are in V
, and construct the two sequences again in the same fashion, with V
being included in the sequence of open sets that we used to define the sequence xk
.
We repeat this. Thus we construct a bounded sequence that does not converge - because otherwise, it would be covered by an open set in the sequence we have choosen.
Hence, if A
is closed and bounded, every open covering has a finite subcover. qed

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\int_{0}^{\infty} e^{-x^2} , dx = \frac{\sqrt{\pi}}{2} ]

[@aninditabanerjee6862]

3
The best proof I've seen is a topological one, using a few theorems about different topologies.
Namely: 1) In an order topology, [a,b]
is a compact set. 2) The order topology on R
is equivalent to the metric induced topology on R
. 3) Closed subsets of compact sets are compact.
Compact subsets of Hausdorff spaces are closed, and in the metric topology we can take Bn(0)
, the ball of radius n
around 0, and since ∪∞n=1Bn(0)=Rm
, we can choose a finite subcover, and so it is a bounded set.
That proves necessity. To prove sufficiency we simply note the following.Let K
be a closed, bounded subset of Rn
. Then K⊆[−M,M]n
is actually compact, as [−M,M]
is compact by 1), and since the order and metric topologies are equivalent, it is compact in the metric topology. By Tychonoff's Theorem(not even necessary, finite products of compact sets are compact), we have [−M,M]n
is also compact. Closed subsets of compact sets are compact, and K
is a closed subset of [−M,M]
, so K
is compact, proving sufficiency.
One can use the following lemma to prove it: Let (Rn,d)
be a Euclidean space with either the Euclidean metric or the taxicab metric. Then every bounded sequence in (Rn,d)
has at least one convergent subsequence.
0
Use the following 2 propositions as step-stones. (1): Sequences in Rn
converge with respect to the sup norm metric, the taxicab metric, or the Euclidean metric if and only if it's component-sequences converge individually in R
. (2): (Heine-Borel Theorem for the line) X⊆R
is closed and bounded if and only if any sequence in X
has a convergent subsequence converging to a point in X
.By contradiction, we can show that ∀1≤i≤n
, the projection of E on the ith
dimension Ei={xi∈R:xi is the ith component of some x∈E}
is a bounded and closed subset of R
.
Take a sequence