Suppose That X is Sequentially Compact and F is Continuous Show That FX is Sequentially Compact

We've arrived at possibly the most confusing notion in topology/analysis. First, we wish to fulfil an earlier promise: to prove that ifC is a closed and bounded subset ofR andf :R →R is continuous, thenf(C) is closed and bounded. [ As a corollary, iff :R →R is continuous, then the imagef([a, b]) is closed and bounded and thus attains a minimum and maximum. We saw that this has huge implications in optimisation problems. ]

To do that, we define the following.

Definition. A topological space X is said to besequentially compact if every sequence has a convergent subsequence.

Thus, a sequentially compact space is one which is so constrained that an infinite sequence must have infinitely many terms clustering around a point.

sequential_compact

For example, R is not sequentially compact since x_n =n clearly has no convergent subsequence. Also,X = (0, 1) is not sequentially compact since x_n = 1/n has no subsequence which converges in X. In the former case, problem occurs because the points keep getting further. In the latter case, the sequence $x_n = 1/n$ really ought to converge to 0, but that limit is not found in the space.

This observation inspires the following result.

Proposition 1. Let (X, d) be ametric space.

If X is sequentially compact, then it's bounded.

If Y is a sequentially compact subspace of X, then Y is closed in X.

Proof.

SupposeX is not bounded; fix $x\in X.$

We claim that the collection ofd(x,y), fory inX, is not bounded; indeed, if it were bounded by B, anyy and z inX would gived(y,z) ≤d(y,x) +d(x,z) ≤ 2B, which is a contradiction.
Pick $x_n \in X, d(x, x_n) > n.$
The resulting sequence is not even Cauchy since for anyn, the set of $d(x_m, x_n)$ for various m is not bounded, so there existsm>n for which $d(x_m, x_n) >1.$

For the second statement, by theorem 1 here, we only need to show any sequence $(x_n)$ inY converging to $a\in X$ must satisfy $a\in Y.$ But by sequential compactness ofY we must have a subsequence $(x_{n_k})$ which converges to an element of Y. Thus $a\in Y$ indeed. ♦

Exercise

Attempt to prove the second statement of proposition 1 for a topological space X. What additional assumption do you need? [ Answer: X has to be Hausdorff. ]

The next result tells us how to construct new sequentially compact spaces out of old ones.

Proposition 2. Let X and Y be topological spaces.

If f:X→Y is continuous and X is sequentially compact, then so is f(X).

X and Y are sequentially compact if and only if X × Y is.

If X is sequentially compact and $Y\subseteq X$ is closed, then Y is sequentially compact.

Proof.

It turns out forR, the converse to proposition 1 is true.

Theorem 3. If X is a closed and bounded subspace ofR ⁿ, then X is sequentially compact.

Proof (Sketchy)

Via scaling, one may assume $X\subseteq [0, \frac 2 3]^n$ , in which case proposition 2 tells us we only need to show $[0, \frac 2 3]^n$ is sequentially compact and for that, it suffices to show X = [0, 2/3] is sequentially compact. For a sequence (x_n ) inX, write out the decimal expansion of eachx_n ; if more than one expression exists, pick any one. Each x_n may be written in the form (0.abcd…).

For each x_n , take the first digit after the decimal point; since it has only 10 possibilities, one of them (0.a…) must occur infinitely many times. Pick an infinite subsequence with first digit =a. Repeat the argument with this new sequence and find another subsequence with the same first two decimal digits (0.ab…). Iteratively, at the k-th step, we get an infinite subsequence with the same first k decimal digits $(0.d_1 d_2 \ldots d_k).$ This gives $r = (0.d_1 d_2 d_3\ldots) \in [0, \frac 2 3].$ Clearly, there's a subsequence of (x_n ) converging to r. Sincex_n ≤ 2/3, we also haver ≤ 2/3. ♦

Together with propositions 1 and 2, we get:

Corollary 4. A subset X ofR is sequentially compact iff it's bounded, and closed in R.

If X is a closed and bounded subset ofR, and f :R →R is continuous, then f(X) is also closed and bounded.

blue-lin

Compact Spaces

We shall prove that for metric spaces, sequential compactness is equivalent to another topological notion.

Definition. A topological space X is said to be compact if

Thus, every open cover of X has a finite subcover, where by "open cover of X", one means a collection of open subsets of X whose union equals X.

compact_spaces

warning Even though "sequential compactness" and "compactness" are both topological concepts which happen to be equivalent for metric spaces, they are not equivalent for general topological spaces. Indeed, the problem lies with our usage of sequences in sequential compactness. If we replace sequences with nets, the result will be true.

Big Theorem. A metric space (X, d) is sequentially compact if and only if it's compact.

Step 1 : Prove that compact implies sequentially compact.

Suppose $(x_n)$ is a sequence of elements in a compact metric space (X,d) which has no convergent subsequence. Fixx inX; there's an open ballN(x, ε(x)) which contains only finitely many terms of the sequence. [ Otherwise, each N(x, ε) contains infinitely many terms of the sequence. Thus we can pickn ₁ < n ₂< n ₃ < … such that $d(x_{n_k}, x) < \frac 1 k$ and $(x_{n_k}) \to x.$ ]

Now the collection of all these open balls is an open cover forX so it has a finite subcover $N(x_i, \epsilon(x_i)),$ which contradicts the fact that each $N(x_i, \epsilon(x_i))$ contains only finitely many terms of the sequence.

Step 2 : IfX is sequentially compact and {U_i } is an open cover ofX, then there is an ε>0 such thatany open ballN(x, ε) is contained in some U_i .

If not, then for ε = 1/n (n = 1, 2, …), some open ballN(x_n , 1/n) is not contained in any U_i . By sequential compactness, we can find a convergent subsequence $(x_{n_k}) \to a.$ Replacing (x_n ) with $(x_{n_k}),$ we may assume that (x_n ) →a andN(x_n , 1/n) is not contained in anyU_i (since 1/n_k < 1/n).

Now thisa lies in someU_i , soN(a, 1/m) lies in U_i for some m. Since (x_n ) →a, there's an indexN such that ifn >N thend(a, x_n ) < 1/(2m). If we pickn > max(N, 2m), thend(a,x_n ) < 1/(2m) gives:

$N(x_n, \frac 1 n) \subseteq N(x_n, \frac 1{2m}) \subseteq N(a, \frac 1 m) \subseteq U_i.$

This contradicts our choice of (x_n ).

Definition. This ε>0 is called aLebesgue number of the open cover {U_i }.

Step 3 : If X is sequentially compact, then for any ε>0, we can coverX with finitely many open ballsN(x,ε).

Suppose this is not true for some ε>0. Let's construct a sequence (x_n ) such that any two distinct terms satisfyd(x_n , x_m ) ≥ ε. Suppose we already have {x ₁,x ₂, …,x_n }. To find the next term, since the union of N(x_i , ε) is not the whole ofX, we can still find x_n+1 outside this union. Thus,d(x_n+1 ,x_i ) ≥ ε fori = 1, 2, …,n. Such a sequence has no convergent subsequence.

Definition. A metric space X in which for any ε>0, X can be covered by finitely many N(x, ε), is said to betotally bounded.

totally_bounded

Step 4 : Prove that sequentially compact implies compact.

If {U_i } is an open cover of X, by step 2, pick ε>0 such that any open ballN(x, ε) is contained in some U_i . By step 3, the whole ofX can be covered by finitely manyN(x_i , ε),i = 1, …,n. EachN(x_i , ε) is contained in some U_i ; and soX can be covered by finitely many U_i . ♦

In the process, we've also proved the following:

Corollary. If (X, d) is a compact metric space, then X is totally bounded and has a Lebesgue number ε>0.

blue-lin

Replacing Sequences with Nets

If we replace the condition "every sequence has a convergent subsequence" with "every net has a subnet", then this is indeed equivalent to the notion of compactness for arbitrary topological spaces.

Before we state the result, though, we should be clear what we mean by "subnets". In the case of sequences, we don't want people to cheat by just taking the first 10 terms. Likewise, in the case of subnets, we force them to contain terms which are indexed by arbitrarily large indices.

Definition. If $(x_i)_{i\in I}$ is a net indexed by directed set I, then a subnet is given by $(x_j)_{j\in J}$ for some subset J of I such that:

for any $i\in I$ , there exists an index j ≥ i contained in J (note that this condition implies J is a directed set).

Now we're ready to state our theorem.

Theorem. A topological space X is compact if and only if every net has a convergent subnet.

Forward Implication.

SupposeX is compact and $(x_i)_{i\in I}$ is a net indexed by directed setI. Assume it has no convergent subnet, so every $a\in X$ is not a limit of some subnet. Thus there's an open subsetU(a) containinga, and an indexi, such thatU(a) does not contain x_j for allj ≥i. [ If not, for each open subsetU containinga and indexi, there's aj ≥i such that $x_j\in U.$ This gives a subnet (x_j ) which converges toa. Contradiction. ]

Since $X = \cup_{a\in X} U(a)$ we can find a finite subset {a ₁,a ₂, …,a_n } of X such that $X = \cup_{k=1}^n U(a_k).$ Also for eachk=1, …,n, there's an indexi_k such thatU(a_k ) does not containx_j for allj ≥ i_k . SinceI is a directed set, we'll pickj ≥ all i_k 's, and so $X = \cup_{k=1}^n U(a_k)$ does not contain x_j which is absurd.

Backward Implication

Suppose every net inX has a convergent subnet and $\{U_j\}_{j\in J}$ is an open cover ofX indexed byJ. We construct a net as follows: the index set isI, the collection of all finite subsets ofJ, ordered by inclusion. Thus, if $S, T\in I,$ then $S\le T \iff S\subseteq T.$ For each $S\in I,$ define the open subset $U_S := \cup_{j\in S} U_j.$ If {U_j } has no finite subcover, then eachU_S ≠X so we can pick $x_S\in X - U_S.$ This gives a net (x_S ).

constructing_net

By assumption, (x_S ) has a convergent subnet (x_T ) →a. Nowa lies in some U_j ; there exists an indexT' such that for allT ≥T', $x_T\in U_j.$ If we pickT containingj, then $x_T \not\in U_T \supseteq U_j$ so $x_T\not\in U_j$ which is a contradiction. ♦

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Source: https://mathstrek.blog/2013/02/24/topology-sequentially-compact-spaces-and-compact-spaces/

Suppose That X is Sequentially Compact and F is Continuous Show That FX is Sequentially Compact

Compact Spaces

Replacing Sequences with Nets

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