Explaining maths

See also  https://math.stackexchange.com/questions/9964/if-no-cauchy-subsequence-exists-must-a-uniformly-separated-subsequence-exist  and  https://math.stackexchange.com/questions/7210/which-metric-spaces-are-totally-bounded  Let's list the conclusions from my recent posts. We'll use the same definitions and notation as before. These results are presumably all well known, though I haven't personally seen explicit statements of all of them before. I had some fun working out the details for myself! Notes on terminology : I have seen both "uniformly separated" and "separated" used for sequences. But there is a completely different notion of separated sets for pairs of sets. So possibly my use of the term "separated set" is confusing. I have seen the term "uniformly discrete" used, but I don't think that I have yet found the most definitive standard terminology. Theorem 1 Let \((X,d)\) be a metric space. Then  the following statements

 Following up on my previous posts about total boundedness and Cauchy sequences, let's look at the special case of countable metric spaces, and see whether any of the proofs become simpler. Along the way we can put together some of the results and arguments from earlier posts about sequences and their subsequences. Of course, finite metric spaces are not very interesting. They are all discrete, compact, sequentially compact, totally bounded and complete. Indeed, every sequence in such a metric space has a constant subsequence. So we will look only at countably infinite metric spaces.  Setting and terminology Throughout this post we work in some countably infinite  metric space \((X,d)\), which we enumerate as \(\qquad X=\{a_1,a_2,a_3,\dots\}\) with all the \(a_n\) being distinct . [This fits with what I think is meant by an enumeration of a set. Not all authors agree: some authors might allow an enumeration to include repeats. Still, it looks like my version agrees with the "d

I posted the following in December 2021 on my first-year pure module's Piazza forum. Hi everyone I saw this a few years ago on QI, and found another version on the web at http://mathandmultimedia.com/2014/12/02/merry-christmas-equation/ If we are told that \(\qquad\displaystyle y=\frac{\ln(\frac{x}m-sa)}{r^2}\) multiplying by \(r^2\) gives \(\qquad yr^2=\ln(\frac{x}m-sa)\,.\) Then taking \(e\) to the power of both sides give \(\qquad\displaystyle e^{yr^2}=\frac{x}m-sa\,,\) and multiplying both sides by \(m\) gives \(\qquad me^{yr^2}=x-msa\,,\) which can be rewritten as \(\qquad me^{rry}=x-mas\,.\) (And a Happy New Year!) Best wishes, Dr Feinstein

 In the last post we were looking at sequences which have no Cauchy subsequences.  Setting and terminology Throughout this post we work in some metric space \((X,d)\). Where we discuss sequences, we mean infinite sequences, and by default our indices will be the positive integers. So (for example), by default, \((x_n)\) will denote a sequence \((x_n)_{n=1}^\infty\). Definition Let \((x_n)\) be a sequence in \(X\). Then \((x_n)\) is uniformly separated if there exists \(\varepsilon>0\) such that for all \(m,n \in \N\) with \(m\neq n\), we have \(\qquad d(x_m,x_n)\geq \varepsilon\,.\) More generally, let \(Y\) be a subset of  \(X\). Then we say that \(Y\) is uniformly separated  if there exists \(\varepsilon>0\) such that for all \(y,y' \in Y\) with \(y\neq y'\), we have \(\qquad d(y,y')\geq \varepsilon\,.\) Here we see that a sequence \((x_n)\) in \(X\) is uniformly separated if and only if all the terms in the sequence are distinct and the set \(\{x_n:n \in \N\}\) is

In the theory of Metric and Topological Spaces, it is standard that compactness and sequential compactness are equivalent for metric spaces. Also, a metric space is compact if and only if it is both complete and totally bounded. I was just thinking about this again, and wondering about conditions which are equivalent to total boundedness for metric spaces (without assuming completeness). I realised (and this is presumably well known) that a metric space \(X\) is totally bounded if and only if the following condition holds: (CS)\(\quad\)Every sequence in \(X\) has a Cauchy subsequence. [I see something about this on the page  https://en.wikipedia.org/wiki/Totally_bounded_space though it looks a bit confusing there, and there are few details.] This is a slightly weakened form of sequential compactness that I haven't seen before myself, but might be useful. I don't know a name for that condition, but it is equivalent to the completion being sequentially compact, which gives one a

In order to formalise the notion of "connected components", we need to know something about when a union of connected sets might be connected. Of course this isn't generally true, e.g. (with the usual topology) \((0,1) \cup (2,3)\) is disconnected, but is a union of two connected sets. On the other the union of two disjoint connected sets can sometimes be connected. For example, \((0,2)=(0,1]\cup(1,2)\) is connected, even though it is a union of two pairwise disjoint connected sets. [Can you work out what the true story is here?] Let's have a look at Lemma 6.12 which gives a sufficient condition for a union of connected sets to be connected.  Informally, these connected pieces are somehow all "attached" to each other at the point \(x_0.\) The proof of Lemma 6.12 makes this formal. There are many ways to prove this lemma. For example, we can assume that the union is equal to \(X\) if we like. We can also use a variety of conditions equivalent to connected