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Let be a compact metric space and let be a sequence of compact sets in . Is it true that is compact?
The answer is no since we can take and . Hence
which is clearly not compact.
For which positive integers does there exist expression
where each is a disk of radius such that each point belongs to either the boundary of some or to precisely interiors of the sets , , ?
(Euler Competition , 2017)
Further, consider any disk and let us show that the number of the other disks it intersects is finite. Indeed, each disk that intersects would have to be a subset of the disk or radius co-centric with . All such disks together would cover an area of at most so that each point is covered by at most disks. Hence the total area of disks that intersect is at most and therefore there are not more than of them.
Now, let us introduce the function whose value is the number of disks that belongs to. Note that
- for almost all
- is constant on every connected component of .
What remains to be shown is that this is impossible. Since the disk intersects only with finitely many other disks , let us consider the circular arc between two such intersection points such that the arc in the figure above. But then obviously the value for points on one side of the arc and for points on the other side of the arc differ by . For the collections of disks that cover the two regions is different – is only on one side of the circular arc.
Let be a compact metric space and be an isometry. Prove that is onto. Use the above result to deduce that Hilbert space is not compact.
[The Hilbert space is the space of all real sequences whose series converges and it is endowed with the metric
It holds that for all . Since is compact it is also sequential compact and thus the sequence has a convergent subsequence. Therefore there exist such such that .
But is an isometry thus
which is clearly an obscurity. Hence is onto.
Well, in we consider the forward shift operator defined as and
which is an isometry but not onto. Hence is not compact.
Prove that there exists no continuous and map ( depiction ) from a sphere to a proper subset of it.
A real number is said to be dyadic rational provided there is an integer and a non negative integer for which . For each and each set:
(i) Prove that the dyadic numbers are dense in .
(ii) Let be the function to which the sequence converges pointwise. Prove that does not exist.
(iii) Show that the convergence is not uniform.
(i) Let such that . We want to show that there exists a dyadic rational in the open interval . By the Archimeadean property we can choose a positive integer such that so that we know that there is an integer such as hence . Now we can easily conclude that the dyadic numbers are dense in .
(ii) If then forall . This says that forall . If is not dyadic rational then forall . Hence the limit of the function is:
Let us now consider a partition of . By the density of the dyadic numbers for each subinterval of the partition. This says that the upper integral of is equal to . Similarly, since the irrationals are dense in it follows that the lower integral of is . Thus the requested integral does not exist.
(iii) Well this pretty much follows from the previous question. Our function is not integrable on and the result follows.