Is the infinite union compact?

Let M be a compact metric space and let A_i be a sequence of compact sets in M. Is it true that \bigcup \limits_{i=1}^{\infty} A_i = A is compact?

Solution

The answer is no since we can take A_i = \left[0,1-\frac{1}{i} \right] and M=[0, 1]. Hence

    \[A = \bigcup_{i=1}^{\infty} \left[0,1-\frac{1}{i} \right] = [0, 1)\]

which is clearly not compact.

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Does there exist an expression?

For which positive integers n does there exist expression

    \[\mathbb{R}^2 = \bigcup_{m=1}^{\infty} A_m\]

where each A_m is a disk of radius 1 such that each point x \in \mathbb{R}^2 belongs to either the boundary of some A_m or to precisely n interiors of the sets A_1 , A_2 , \dots ?

(Euler Competition , 2017)

Solution [official]

For none n. Assume , on the contrary , that such expression exists for some value of n. Let us first note that the set of disks is countable hence the Lebesgue measure of the union of all the circles is 0.

Further, consider any disk A_i and let us show that the number of the other disks it intersects is finite. Indeed, each disk that intersects A_k would have to be a subset of the disk or radius 3 co-centric with A_k. All such disks together would cover an area of at most 9 \pi so that each point is covered by at most n disks. Hence the total area of disks that intersect A_k is at most 9 \pi n and therefore there are not more than 9n of them.

Now, let us introduce the function f:\mathbb{R}^2 \rightarrow \mathbb{N} whose value f(P) is the number of disks A_i that P belongs to. Note that

  • f(P)=n for almost all P \in \mathbb{R}^2
  • f(P) is constant on every connected component of \mathbb{R}^2 \setminus \bigcup \limits_{k=1}^{\infty} \partial A_k.

What remains to be shown is that this is impossible. Since the disk A_i intersects only with finitely many other disks , let us consider the circular arc between two such intersection points such that the arc XY in the figure above. But then obviously the value f(P) for points on one side of the arc and f(P) for points on the other side of the arc differ by 1. For the collections of disks that cover the two regions is different – A_i is only on one side of the circular arc.

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Isometries in a compact space

Let (X, \rho) be a compact metric space and f:X \rightarrow X be an isometry. Prove that f is onto. Use the above result to deduce that \ell^2 Hilbert space is not compact.

[The \ell^2 Hilbert space is the space of all real sequences x_n whose series \sum \limits_{n=1}^{\infty} x_n^2 converges and it is endowed with the metric

    \[\rho \left ( x_n , y_n \right ) = \sqrt{\sum_{n=1}^{\infty} \left ( x_n - y_n \right )^2}\]

Solution

Suppose that f is not onto. Then , there exists an x \in X such that x \notin f(X). Since f is an isometry it is also continuous and because X is compact f(X) is also a compact subset of X. Hence a=\rho(X, f(X))>0. Consider the sequence

    \[x_0=x \;, \; x_1=f(x_0) \; , \cdots \;, \; x_{n} = f\left ( x_{n-1} \right )\]

It holds that x_n \in f(X) for all n \geq 1. Since f(X) is compact it is also sequential compact and thus the sequence \{x_n\}_{n \geq 1} has a convergent subsequence. Therefore there exist m, n such m>n \geq 1 such that \rho(x_m , x_n)<a.

But f is an isometry thus

\rho (x_m, x_n)=\rho (f(x_{m-1}), f(x_{n-1}))= \cdots = \rho (x_{m-n}, x)

However, x_{m-n} \in f(X) hence

    \[\rho(x,f(X))\leq \rho(x,x_{m-n})=\rho (x_m, x_n) < a\]

which is clearly an obscurity. Hence f is onto.

Well, in \ell^2 we consider the forward shift operator defined as S_r:\ell^2 \rightarrow \ell^2 and

    \[S_r(x_1, x_2, x_3, \cdots ) = (0,x_1, x_2, \cdots)\]

which is an isometry but not onto. Hence \ell^2 is not compact.

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No map exists

Prove that there exists no continuous and 1-1 map ( depiction ) from a sphere to a proper subset of it.

Solution

Let x_0 \in \mathbb{S}_n and consider f:\mathbb{S}_n \rightarrow \mathbb{S}_n \setminus \{x_0\} that is a continous and 1-1 map . We consider the stereographic projection g:\mathbb{S}_n \setminus \{x_0\} \rightarrow \mathbb{R}^n that is also 1-1 and continous. Hence the composition \left( g \circ f \right): \mathbb{S}_n \rightarrow \mathbb{R}^n is continuous and 1-1  which is an obscurity due to Borsuk-Ulam.

The exercise can also be found at mathematica.gr .

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On dyadic numbers

A real number x is said to be dyadic rational provided there is an integer k and a non negative integer n for which \displaystyle x=\frac{k}{2^n} . For each x \in [0, 1] and each n \in \mathbb{N} set:

f_n(x) = \left\{\begin{matrix} 1 &, & x =\dfrac{k}{2^n} , \; k \in \mathbb{N} \\\\ 0& , & \text{otherwise} \end{matrix}\right.

(i) Prove that the dyadic numbers are dense in \mathbb{R}.

(ii) Let f:[0, 1] \rightarrow \mathbb{R} be the function to which the sequence \{f_n\}_{n \in \mathbb{N}} converges pointwise. Prove that \bigintsss_0^1 f(x) \, {\rm d}x does not exist.

(iii) Show that the convergence f_n \rightarrow f is not uniform.

Solution

(i) Let a, b \in \mathbb{R} such that a<b . We want to show that there exists a dyadic rational in the open interval (a, b). By the Archimeadean property we can choose a positive integer such that 2^n (b-a) >1 so that we know that there is an integer k such as 2^n a <k <2^n b hence \displaystyle a<\frac{k}{2^n}< b. Now we can easily conclude that the dyadic numbers are dense in \mathbb{R} .

(ii) If  \displaystyle x=\frac{k}{2^n} \in [0, 1] then \displaystyle x=\frac{k 2^r}{2^{n+r}} forall  r>0 . This says that f_m(x) =1 forall m>n. If x is not dyadic rational then f_n(x)=0 forall n \in \mathbb{N} . Hence the limit of the function is:

f(x)= \left\{\begin{matrix} 1 &, & x \; \text{is a non zero dyadic rational} \\\\ 0& , & \text{otherwise} \end{matrix}\right.

Let us now consider a partition \mathcal{P}=\{x_0, x_1, \dots, x_n \} of [0, 1] . By the density of the dyadic numbers \sup \left \{ f(x): x \in [x_{i-1}, x_i] \right \}=1 for each subinterval of the partition. This says that the upper integral of f is equal to 1. Similarly, since the irrationals are dense in [0, 1] it follows that the lower integral of f is 0. Thus the requested integral does not exist.

(iii) Well this pretty much follows from the previous question. Our function is not integrable on [0, 1] and the result follows.

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