(last updated: 2:25:18 PM, October 08, 2020)
Definition 2.5.1 Let \(\epsilon>0\) be given. By a neighborhood for a real number \(s\), we mean a set \(N_\epsilon(s)=\{x\mid|x-s|<\epsilon\}\). A deleted (or punctured) neighborhood of \(s\) is a set \(N_\epsilon^{-}(s)=\{x\mid0<|x-s|<\epsilon\}\).
It is important to understand there are two different neighborhoods around a point, the one that includes it and the one that excludes it. Think of a line segment with the centerpoint punched out (deleted) vs the whole line segment. Generally neighborhoods have open ends, not inclusive of the endpoints like \((a,b)\).
“Neighborhoods” are a common concept in analysis courses. Often you wish to construct an interval, disk, or “ball” around a point. These neighborhoods don’t have to have any specific geometrical interpretation or shape though. The concept of distance can even be generalized; there are ways to measure distance that is not the conveption you encountered in calculus.
Definition 2.5.2 Let \(S\) be a set of real numbers. A real number \(s_0\) is an accumulation point of \(S\) if and only if for any \(\epsilon>0\), there exists at least one point \(t\) of \(S\) such that \(0<|t-s_0|<\epsilon\).
I like to envision the accumulation point \(a\) as a tiny attractive particle with the other numbers in the set being like little point particles clustering around \(a\).
Note that:
Example: A finite set has no accumulation points.
For a concrete example, consider \(S=\{1,2,3\}\). Let \(\epsilon=1\). Then \((1-\epsilon,1+\epsilon)=(0,2)\) only contains \(1\) and no other points of \(S\). The same is true for \(2\) and \(3\). SO none of these are accumulation points.
The general case: Consider \(S=\{s_1,s_2,\ldots,s_n\}\). Then we can take the maximal distance between any of these elements: \(d=\max\{|s_i-s_j\mid i,j\leq n\}\). We know this is strictly positive since \(s_i\neq s_j\) whenever \(i\neq j\). This is because you only list a number once in a set, i.e. \(\{3,3,5\}\) is not a set, but \(\{3,5\}\) is. Now choose any \(\epsilon\) that satisfies \(0<\epsilon<d\). Then for some fixed but arbitrary \(j\), \(s_i\in(s_j-\epsilon,s_j+\epsilon)\) if and only if \(i=j\). Thus we have found an \(\epsilon\)-neighborhood of \(s_j\) that only contains \(s_j\) and no other points of \(S\). Hence \(s_j\) is not an accumulation point.
Example: Consider the set \(S=\{\frac1n\mid n\in\mathbb N\}\). Then \(0\) is an accumulation point of this set. For any \(\epsilon>0\) there are infinitely many \(n\) such that \(\frac1n\in(0,\epsilon)\). Note that \(0\not\in S\).
If we were to include \(0\) in \(S\) by \(S_0=S\cup\{0\}\), then \(0\) is still the only accumulation point of \(S_0\).
Example: Consider the set \(S=\mathbb Q\cap (0,1)\) be the set of all rational numbers between 0 and 1 (not including the endpoints 0 and 1). Then every real number in the entire closed real interval \([0,1]\) is an accumulation point of \(S\). For any \(\epsilon>0\) and any \(x\in[0,1]\) there are infinitely many rational numbers in the part of the deleted neighborhood of \(x\) that overlaps with \(S\). This could look like \((x-\epsilon,x+\epsilon)\), \((0,x+\epsilon)\), or \((x-\epsilon,1)\). This is due to the “density of rationals” (see Theorem 1.7.10).
This next result is a statement about the structure of the real line. We are allowed to have an infinite quantity of points packed into a finite length interval. This means that those points need to be clustered in at least one location. They may be clustered around many locations as well.
Theorem 2.5.4 (Bolzano-Weierstrass Theorem for Sets) Every bounded, infinite set of real numbers has at least one accumulation point.
Proof. Consider a bounded set of infinitely-many real numbers \(S\). This set is bounded from above and below, therefore by completeness it has a finite supremum \(\sup S\) and infemum \(\inf S\) that exist. Let \(a_1=\inf S\) and \(b_1=\sup S\) and consider the interval \(I_1=[a_1,b_1]\). We know that every element of \(S\) is in this interval. This interval has length \(b_1-a_1\).
Now split this interval in two: \([a_1,(a_1+b_1)/2]\) and \([(a_1+b_1)/2,b_1]\). Since \(S\) has infinitely many numbers in it, at least one of these subintervals has infinitely many numbers. Chose that subinterval. If both have infinitely many numbers, then just choose either one. We just need each subinterval in this process to still contain infinitely many of the numbers in \(S\). Call this subinterval \(I_2=[a_2,b_2]\). Note that either \(a_2=a_1\) or \(b_2=b_1\) is true (not both though), and the other endpoint is \((a_1+b_1)/2\), the midpoint of our starting interval. This subinterval has length \((b_1-a_1)/2\).
Now again split the preeceeding subinterval in two: \([a_2,(a_2+b_2)/2]\) and \([(a_2+b_2)/2,b_2]\) and note that at least one of these intervals must still contain infinitely many elements of \(S\). Choose this next subinterval so that it contains infinitely many elements of \(S\). Call this sub interval \(I_3=[a_3,b_3]\) and note that it has length \((b_1-a_1)/4\).
Continue in this way constructing intervals \(I_1,I_2,I_3,\ldots\). Eaach \(I_k\) will have length \(b_k-a_k=\frac{(b_1-a_1)}{2^k}\). We have also created two sequences \(\{a_n\}\) and \(\{b_n\}\). Note that: \[a_1\leq a_2 \leq \cdots \leq a_k \leq a_{k+1} \leq \cdots \leq b_{k+1} \leq b_k \leq \cdots \leq b_2 \leq b_1\] so that \(a_n\) is increasing and bounded from above by \(b_1\) and \(b_n\) is decreasing and bounded form below by \(a_1\), thus \(a_n\) and \(b_n\) converge. Let \(a_n\rightarrow A\) and \(b_n\rightarrow B\). Note that \(A\leq B\) by comparison. In fact \(a_k\leq A\leq B\leq b_k\) for any \(k\in\mathbb N\). This tells us that \(B-A\leq b_k-a_k=\frac{b_1-a_1}{2^k}\) for all \(k\). For any \(\epsilon>0\) we can find a \(k\) such that \(\frac{b_1-a_1}{2^k}<\epsilon\) thus we must have that \(A=B\).
Also note that each \([a_k,b_k]\) has infinitely many elements of \(S\) in it. Therefore, for any \(\epsilon>0\), we can find a \(k\) such that \(A\in [a_k,b_k]\) and \(b_k-a_k<\epsilon\). So we have \(a_k\leq A\leq b_k\) and thus \(b_k-A<\epsilon\), \(A-a_k<\epsilon\). This shows that \(A-\epsilon<a_k\leq b_k\leq A+\epsilon\). Therefore there are infinitely many elements of \(S\) (distinct form \(A\)) in the interval \((A-\epsilon, A+\epsilon)\). Thus \(A\) is an accumulation point of \(S\).
Note that it is a fact that \(S\) contains infinitely many distinct numbers. This means that at least one of our sequences \(a_n\) or \(b_n\) is definitely not eventually constant and therefore each \(I_k\) interval does indeed contain infinitely many distinct elements of \(S\). This is important because being an accumulation point means that the deleted neighborhood contains infinitely many points of the set. \(\blacksquare\)
Definition 2.5.6 A sequence \(a_n\) is called a Cauchy sequence if and only if for each \(\epsilon>0\) there exists \(n^*\in\mathbb N\) such that \(|a_n-a_m|<\epsilon\), for all \(n,m\geq n^*\).
This definition looks very much like the definition of convergence, but it is slightly different. A Cauchy sequence has its terms getting very close together eventually, but we are not necessarily given that the sequence converges. In \(\mathbb R\), a Cauchy sequence does indeed converge to a real number as we will see shortly, but in other spaces Cauchy sequences may not converge. For example, \(\mathbb Q\) is an ordered field, and a Cauchy sequence of rational numbers need not converge (to a rational number). It will converge to an irrational number in that case. The property of “convergence” depends on converging to an object that exists within the overall parent set or universe that we are considering. Our universe is the set of real numbers. And it is indeed a fat that every Cauchy sequence of real numbers converges toa real number.
Example: Show that \(a_n=\frac1n\) is Cauchy.
We want \(|\frac1n-\frac1m|<\epsilon\). We can compute that \(|\frac1n-\frac1m|=\frac{|m-n|}{nm}\leq \frac{n+m}{nm}=\frac1n+\frac1m\). So let’s just choose \(n^*>\frac2\epsilon\) so that \(n,m\geq n^*\) implies that \(\frac1n,\frac1m <\frac\epsilon2\). This gives \(|\frac1n-\frac1m|\leq \frac1n+\frac1m<\frac\epsilon2+\frac\epsilon2=\epsilon\).
Some common mistakes: \(|a_n-a_{n+1}|<\epsilon\) evnetually does not imply the sequence is Cauchy. Similarly, showing any two terms separated by a fixed index-distance get closer over time does not imply the sequence is Cauchy, i.e. \(|a_n-a_{n+k}|<\epsilon\) for some fixed \(k\) does not imply we have a Cauchy sequence. COnsider \(a_n=(-1)^n\) and notice that \(|a_n-a_{n+2}|=0\) for all \(n\). This sequence is definitely not Cauchy though since for any \(n^*\) there are odd and even \(n,m\geq n^*\) such that \(|a_n-a_m|=2\) which will be greater than any \(\epsilon<2\).
Theorem 2.5.8 Every Cauchy sequence is bounded.
Proof. We want \(|a_n|\leq M\) for all \(n\). We know that given some \(\epsilon>0\) there is a cutoff \(n^*\) such that \(n,m\geq n^*\) implies \(|a_n-a_m|<\epsilon\). In particular \(|a_{n^*}-a_m|<\epsilon\) for all \(m\geq n^*\). Thus \(a_{n^*}-\epsilon<a_m<a_{n^*}+\epsilon\) so that the tail of the sequence \(\{a_m\}_{m\geq n^*}\) is bounded. In particular, \(|a_n|\leq |a_{n^*}|+\epsilon\) for \(n\geq n^*\). Of course, the first part of the sequence is a finite list of numbers and so is also bounded. More specifically \(|a_n|\leq \max_{k<n^*}\{|a_k|\}\) for all \(n<n^*\). Now let \(M=\max\{|a_1|,|a_2|,\ldots,|a_{n^*-1}|,|a_{n^*}|+\epsilon\}\). Then \(|a_n|\leq M\) for all \(n\). \(\blacksquare\)
Theorem 2.5.9 In \(\mathbb R\), a sequence is Cauchy if and only if it is convergent.
Proof. Assume the sequence is convergent, \(a_n\rightarrow A\). Now choose a cutoff \(n^*\) such that \(n\geq n^*\) implies that \(|a_n- A|<\frac\epsilon2\). Then for any \(n,m\geq n^*\) we have that \(|a_n-a_m|=|a_n-A+A-a_m|\leq |a_n-A|+|A-a_m|<\frac\epsilon2+\frac\epsilon2\).
Now assume the sequence is Cauchy. Consider the set \(S=\{a_n\mid n\in\mathbb N\}\). There are two possibilities: either \(S\) has a finite number of points or an infinite number of points.
If this set has a finite number of points, then you know there is a number \(A\) such that \(a_n=A\) for infinitely many \(n\). Thus for any \(n^*\), there are infinitely many \(m\geq n^*\) such that \(a_m=A\). Then I claim \(a_n\) converges to \(A\). To see this, fix \(\epsilon>0\) and choose \(n^*\) such that \(n,m\geq n^*\) implies that \(|a_n-a_m|<\epsilon\) (we can do this because we assume the sequence is Cauchy). Now just chose any arbitrary \(m\geq n^*\) such that \(a_m=A\). But we must then have that \(|a_n-A|=|a_n-a_m|<\epsilon\) for all \(n\geq n^*\). Thus \(a_n\) converges to \(A\).
If \(S\) has an infinite number of points, then we use the fact that every cauchy sequence is bounded, hence set \(S\) is bounded, and therefor has at least one accumulation point (Bolzano-Weierstrass Theorem 2.5.4). Let \(A\) be an accumulation point. This means that for any \(\epsilon>0\) there are infinitely many \(m\) such that \(|a_m-A|<\epsilon\). Now choose a cutoff \(n^*\) such that \(n,m\geq n^*\) implies that \(|a_n-a_m|<\frac\epsilon2\) (since the sequence is assumed to be Cauchy). Now when \(n\geq n^*\) we can just pick some arbitrary \(m\geq n^*\) such that \(|a_m-A|<\frac\epsilon2\) (since \(A\) is an accumulation point of \(S\)). This gives us that \(|a_n-A|=|a_n-a_m+a_m-A|\leq |a_n-a_m|+|a_m-A|<\frac\epsilon2+\frac\epsilon2=\epsilon\). Thus \(a_n\) converges to \(A\). \(\blacksquare\)
It is important here that \(\mathbb R\) is complete. There are other types of sets/spaces that are not complete but still have Cauchy sequences that are not convergent. In fact \(\mathbb Q\) by itself still has Cauchy sequences that do not converge (the converge to an irrational number).
There are many other “exotic” mathematical spaces that are complete and many that are not complete. In teh complete spaces, Cauchy sequences always converge to an element in the space. Int eh spaces that are not complete, there can still be Cauchy sequences, but they don’t converge. You can think of it as their converging to something outside of the space under consideration. Often enlarging the space to “complete” it will allow Cauchy sequences to thus converge to an element in the space.
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