10.24 Absolute value and Supremum

TutorialsWorshopsHomework 4.pdf

1. 1. Assume \(s \in \mathbb{R}\) is an upper bound for a set \(A \subseteq \mathbb{R}\). Then \(s = \sup A\) if and only if, for every choice of \(\varepsilon > 0\), there exists an element \(a \in A\) satisfying \(s - \varepsilon < a\).

      Proof

      Since \(A\neq \emptyset\) and \(s\) is an upper bound for \(A\), then \(A\) has a supremum.

      \(\Rightarrow\)) Goal: \(s=supA\Rightarrow \forall \varepsilon>0,\exists a\in A\) such that \(s-\varepsilon<a\).

      Suppose \(s-\varepsilon\geq a\), then \(s-\varepsilon\) is an upper bound.

      Since \(s=supA\), then \(s<s-\varepsilon\). Contradiction!

      \(\Leftarrow\)) Goal: \(\forall \varepsilon>0,\exists a\in A\) such that \(s-\varepsilon<a\Rightarrow s=supA\)

      Let \(b\) another upper bound for \(A\), we need to prove that \(s\leq b\)

      Suppose that \(s>b\Rightarrow s-b>0\), then take \(\varepsilon=s-b\) by hypothesis, \(\exists a\in A\) such that \(s-\varepsilon<a\leq s\Rightarrow s-(s-b)<a\leq s\Rightarrow b<a\leq s\) Contradiction. Then \(s\leq b\)​ 2. Assume \(i \in \mathbb{R}\) is a lower bound for a set \(A \subseteq \mathbb{R}\). Then \(i = \inf A\) if and only if, for every choice of \(\varepsilon > 0\), there exists an element \(a \in A\) satisfying \(a < \varepsilon + i\).

      Since \(A\neq \emptyset\) and has a lower bound, then it has an infimum

      \(\Rightarrow\)) Goal: \(i = \inf A\) \(\Rightarrow\) \(\forall \varepsilon > 0\), \(\exists a \in A\) such that \(a < \varepsilon + i\).

      Suppose \(a\geq\varepsilon+i\), then \(\varepsilon+i\) is a lower bound.

      Then \(\varepsilon+i<i\) which is contradiction!

      Thus \(\forall \varepsilon > 0\), \(\exists a \in A\) such that \(a < \varepsilon + i\).

      \(\Leftarrow\)) Goal: \(\forall \varepsilon > 0\), \(\exists a \in A\) such that \(a < \varepsilon + i\) \(\Rightarrow\) \(i = \inf A\)

      We need to prove \(\forall b\in\)​\(LBs\), \(b<i\leq a\)​

      Suppose \(b>i\Rightarrow b-i>0\), then take \(\varepsilon=b-i\Rightarrow b>a\), contradiction!

      Thus \(\forall b\in\)​\(LBs\), \(b<i\leq a\). Thus \(inf=i\)

2. Show that if \(\sup A < \sup B\), then there exists \(b \in B\) that is an upper bound of \(A\).

   Goal: \(\sup A < \sup B\Rightarrow \exists b\in B,b\geq a,\forall a\in A\)

   Suppose \(\forall b\in B,\exists a\in A,a>b\)​

   Then for \(supB\), \(\exists a \in A\) such that \(a>supB\) which is contradiction to the condition!

3. For each of the following sets, compute the supremum and infimum. Justify your answers.

   1. \(\left\{ \frac{1}{n} : n \in \mathbb{N} \right\}\)

      The sequence is increasing such as \(\{1,\frac12,\frac13,\ldots\}\)​

      \(1\) is a upper bound. Since the set is not empty, then it has an supremum. Thus \(sup=1\)

      Then we take \(\forall n\in \N,n>n_0>\frac{1}{\varepsilon}\), then \(|\frac1n-0|<\varepsilon\), thus it converges to \(0\)​

      Then \(inf=0\) 2. \(\left\{ \frac{n}{2n+1} : n \in \mathbb{N} \right\}\)

      \(\frac{n}{2n+1}=\frac{2n}{2\left(2n+1\right)}=\frac{2n+1-1}{2\left(2n+1\right)}=\frac12-\frac{1}{2\left(2n+1\right)}\)​

      Notice: \(2(2n+1)<2(2(n+1)+1)\Rightarrow\frac{1}{2(2n+1)}>\frac{1}{2(2(n+1)+1)}\Rightarrow-\frac{1}{2(2n+1)}<-\frac{1}{2(2(n+1)+1)}\Rightarrow\frac12-\frac{1}{2(2n+1)}<\frac12-\frac{1}{2(2(n+1)+1)}\)​

      Thus the sequence is increasing such as \(\{\frac13,\frac25,\ldots\}\)​

      Then we can see that \(\frac13\) is a lower bound. Moreover since the set is not empty, then it has an infimum.

      Then \(inf=\frac13\)​

      When \(n\) becomes bigger, then \(\frac{1}{2\left(2n+1\right)}\rightarrow0\)​

      Then \(\frac12\) is an upper bound.

      Since \(\frac12-\frac{1}{2(2n+1)}<\frac12\) not in the set and the set is not empty and has a upper bound, then it has a supremum

      Let \(\varepsilon>0\) by Archimedes property, \(\exists N\in \N\) such that \(\frac1N<\varepsilon\).

      Since \(N<2(2N+1)\), then \(\varepsilon>\frac{1}{N}>\frac{1}{2\left(2N+1\right)}\Rightarrow\frac12-\frac{1}{2(2N+1)}>\frac12-\varepsilon\Leftrightarrow\frac12-\varepsilon<\frac12-\frac{1}{2(2N+1)}\)​

      Using exercise 1.a, \(sup=\frac12\)​ 3. \(\{x \in \mathbb{Q} : x > 0 \text{ and } x^2 < 2\}\)

      The set is equivalent to \(\{x\in\mathbb{Q}:0<x<\sqrt2\}\) which can be proved by exercise 3.4 4. Let \(\alpha\) and \(\beta\) be real numbers such that \(\alpha < \beta\). \(\{x \in \mathbb{Q} : \alpha < x < \beta\}\)

      Since the density of \(\mathbb{Q}\) in \(\R\), the set is not empty.

      From the set we know \(\alpha\) is a lower bound and \(\beta\) is an upper bound, then it has a supremum and infimum.

      First, we prove \(sup=\beta\), we need to find \(\beta-\varepsilon<q<\beta,q\in set\)

      Let's analyze: Since \(\beta-\varepsilon\in\mathbb{R}\), by density there must exist such \(q\in set\)

      However, recall that \(\varepsilon\) is arbitrary and \(q\) must in the set, then \(\beta-\varepsilon\) may less than \(\alpha\)

      Then \(q\) may not in the set, thus in this situation we need to take \(q\) such that \(\alpha<q<\beta\)

      Thus we can choose \(\max\{\alpha,\beta-\varepsilon\}\) (\(\beta-\varepsilon\) 有可能在中间或是\(\alpha\)左边) to ensure \(q\) is always in the set

      Which is there exists \(\max\{\alpha,\beta-\varepsilon\}<q<\beta\) such that \(\beta-\varepsilon<q\)

      Now the infimum

      similarly...

      ​ 5. \(\{x \in \mathbb{Q} : -\frac{3}{4} \leq x \leq 0\}\)

      Obviously, the set is not empty, \(0\) is an upper bound and \(-\frac34\) is a lower bound, thus exists sup and inf

      Since \(0\) is in the set and \(-\frac34\) is in the set, then they are sup and inf by theorem

4. Recall that a set \(A\) is dense in \(\mathbb{R}\) if an element of \(A\) can be found between any two real numbers \(a < b\). Which of the following sets are dense in \(\mathbb{R}\)? Take \(p \in \mathbb{Z}\) and \(q \in \mathbb{N}\) in every case.

   Recall: Density of Q in R

   1. The set of all rational numbers \(\frac{p}{q}\) with \(q\) a power of 2.

      \(\left\lbrace\frac{p}{q}:p\in\mathbb{Z},q=2^{i},i\in\mathbb{N}\cup\left\lbrace0\right\rbrace\right\rbrace\)

      First, let \(i\in\mathbb{N}_0\) such that \(\frac{1}{2^{i}}<y-x\) (Archimedes property 2)

      Let \(p\) be the first \(\N\) such that \(p>2^{i}x\Rightarrow\frac{p}{2^{i}}>x\) (Archimedes property 1)

      Let \(q=\frac{p}{2^{i}}\). We have \(x<q\)

      Let \(p-1\leq2^{i}x\) since \(p\) is the first element (WOP) (Trick)

      Then, \(\frac{p-1}{2^{i}}\leq x\)

      Since \(\frac{1}{2^{i}}<y-x\), then \(\frac{p-1}{2^{i}}+\frac{1}{2^{i}}<y-x+x=y\) (\(q=\frac{p}{2^{i}}\))

      Summary

      We can deal with the proof when we meet to variables, since prove the density of Q in R is also two variables 2. The set of all rational numbers \(\frac{p}{q}\) with \(q \leq 10\).

      Hint: \(\frac{p}5\in\) set \(\Rightarrow\frac{1\cdot2\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot10\cdot p}{1\cdot2\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot10\cdot5}=\frac{k}{10!}\) also \(\frac{p}{2}=\frac{L}{10!}\)​

      Thus \(B=\{\frac{p}{q}:p\in\mathbb{Z},q\in\mathbb{N},q\leq10\}\subset\left\lbrace\frac{p}{10!}:p\in\mathbb{Z}\right\rbrace\)​

      Proof

      We will show that \(\left\lbrace\frac{p}{10!}:p\in\mathbb{Z}\right\rbrace\) is not dense in \(\R\)​

      ​

      There is no element of this set between \(a\) and \(b\).

      Then \(\left\lbrace\frac{p}{10!}:p\in\mathbb{Z}\right\rbrace\) is not dense in \(\R\)​

      Also, this can be proved by the similar proof of density of Q in R, but we can find that since \(q\) is limited, we cannot let \(y-x\) too small, thus it is not dense 3. The set of all rational numbers \(\frac{p}{q}\) with \(10|p| \geq q\).

      The set is \(\{\frac{p}{q}\in\mathbb{R}:10|p|\geq q\}\Rightarrow\{\frac{p}{q}\in\mathbb{R}:\frac{|p|}{q}\geq\frac{1}{10}\}\)​

      No!!

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