Mixing
Proof verification for limit point, lim sup and lim inf
Proposition: Let $(a_n)_{n=m}^{infty}$ be a sequence of real numbers, let $L^+$ be the limit superior of the sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+, L^-$ are extened real numbers).
(a) For every $x> L^+$, there exists an $Nge m $ such that $,a_n<x$ for all $nge N$ (the elements of the sequence are eventually less than $x$). Similarly, for every $y<L^-$ there exists an $Nge m $ such that $,a_n>y$ for all $nge N$.
(b) For every $x< L^+$, and every $Nge m$, there exists an $nge N$ such that $a_n> x$ (the elements of the sequence exceed $x$ infinitely often). Similarly for $y>L^-$, and every $Nge m$, there exists an $nge N$ such that $a_n< y$.
(c) We have $,text{inf}_{nge m}(a_n)le L^- le L^+le text{sup}_{nge m}(a_n)$
(d) If $c$ is any limit point of $(a_n)$ then we have $L^- le c le L^+$
(e) If $L^+$ is finite then it is a limit point of the sequence. Similarly if $L^-$ is finite.
(f) Let $c$ be a real number. If $(a_n)rightarrow c$, then we must have $L^+=c=L^-$. Conversely if $L^+=c=L^-$, then $(a_n)rightarrow c$.
I'd like to know if my proof is sound. I'd appreciate any suggestion.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$. Thus by definition this means, $text{inf}_{Nge m} (a_N^{+}) <x$, so $L+varepsilon$ cannot be the infimum of the sequence; and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
(b) Now if $x < L^+=text{inf}_{Nge m} (a_N^{+})le a_N^{+}$. Let us fix some $Nge m$. Thus, $x <a_N^{+}$, and by definition there must be some $nge N$ such that $x< a_n$ (otherwise $x$ would be an upper bound which is less than the least upper bound, which clearly is a contradiction ), as desired. The other part is proven similarly.
(c) Since $text{sup}_{nge m} (a_n) = a_m^+$ and $text{inf}_{nge m} (a_n) = a_m^-$. Clearly, $a_m^-letext{sup}_{Nge m}(a_N^{-})= L^-$ and $L^+=text{inf}_{Nge m} (a_N^{+})le a_m^+$. So, we need to show $L^-le L^+$. Since we have $a_N^{-}le a_N^{+}$ for every $N$. Thus $L^-le L^+$ (because the limit conserve the non-strict inequality), as desired.
(d) Suppose both $L^{+}, L^{-}$ are finite and we let $varepsilon>0$ be given. By (a) there is some $N$ such that $a_n le L^+ +varepsilon$ for each $nge N$. Now, using the definition of limit point, there is some $n_0ge N$ such that $|a_{n_0}-c| le varepsilon$. Thus $,c-L^{+}= (c-a_{n_0})+(a_{n_0}-L^{+})le 2 varepsilon$, since $varepsilon$ was arbitrary, this implies $cle L^{+}$. Similarly, there is some $Mge m$, so that $L^{-}-varepsilon < a_n$ for each $nge M$. By hypothesis $c$ is a limit point and then, for this $M$ there is an $n'ge M$ for which $|a_{n'}-c|le varepsilon$. Thus $L^{-}-c = L^{-} - a_{n'}+a_{n'}-cle 2 varepsilon$, i.e., $L^{-}-cle 2 varepsilon$ for any $varepsilon>0$ and hence $L^{-}le c$.
If $L^{+} = +infty$, then $cle L^{+}$. Now either $L^{-}=+infty$ or $L^{-}=-infty$. The second case is trivial. But in the first case this would imply that the limit point is infinite and we are not defined yet infinite limit points but follows in the exact same way.
(e) Suppose $L^{+}$ is finite and let $varepsilon>0$ be given. Thus, as a consecuence of (a) there is some $N>m$ so that $a_n<L^{+}+varepsilon$ for any $nge N$. Using (b) we know that there at least one $n'ge N$ such that $L^{+}-varepsilon<a_{n'}$. So, $L^{+}-varepsilon< a_{n'}<L^{+}+varepsilon$, i.e., $|a_{n'}-L^{+}|le varepsilon$. Since $varepsilon$ was arbitrary and we can choose any $Nge m$, thus $L^{+}$ is a limit point. The other case is symmetric.
(f) Suppose $(a_n)rightarrow c$; we wish to show $L^{+}= L^{-}$, by (c) we already know that $L^{-}le cle L^{+}$ so, it will suffice to prove $L^{+}le c le L^{-}$.
Let $varepsilon>0$ be given. Since $(a_n)$ converges to a real number, then is bounded. So, the limit superior and inferior are finite and limit points. There is a $Nge m$ such that $|a_n-c|le varepsilon,$ for each $n$, and there is $n'ge N$ so that $L^{+}-varepsilon< a_n'$. Thus $L^{+}-c = (L^{+}-a_n')+(a_n'-c)le 2varepsilon$. And thus $L^{+}le c$. A symmetric argument prove that $L^{-}ge c$ and we're done.
To the other direction we assume that $c=L^{+}=L^{-}$. So, there is some $N_1$ such that $a_n < L^{+} + varepsilon$ and similarly there is some $N_2$ so that $L^{-} - varepsilon< a_n$. We pick the greater between $N_1$ and $N_2$ so, the two inequalities occurs simultaneously. Thus $L^{-} - varepsilon<a_n < L^{+} + varepsilon$ and since both are the same, this would imply $|a_n - c|le varepsilon$ for every $nge N$, which shows that the sequence converges to $c$. $square$
real-analysis proof-verification
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
add a comment |
Proposition: Let $(a_n)_{n=m}^{infty}$ be a sequence of real numbers, let $L^+$ be the limit superior of the sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+, L^-$ are extened real numbers).
(a) For every $x> L^+$, there exists an $Nge m $ such that $,a_n<x$ for all $nge N$ (the elements of the sequence are eventually less than $x$). Similarly, for every $y<L^-$ there exists an $Nge m $ such that $,a_n>y$ for all $nge N$.
(b) For every $x< L^+$, and every $Nge m$, there exists an $nge N$ such that $a_n> x$ (the elements of the sequence exceed $x$ infinitely often). Similarly for $y>L^-$, and every $Nge m$, there exists an $nge N$ such that $a_n< y$.
(c) We have $,text{inf}_{nge m}(a_n)le L^- le L^+le text{sup}_{nge m}(a_n)$
(d) If $c$ is any limit point of $(a_n)$ then we have $L^- le c le L^+$
(e) If $L^+$ is finite then it is a limit point of the sequence. Similarly if $L^-$ is finite.
(f) Let $c$ be a real number. If $(a_n)rightarrow c$, then we must have $L^+=c=L^-$. Conversely if $L^+=c=L^-$, then $(a_n)rightarrow c$.
I'd like to know if my proof is sound. I'd appreciate any suggestion.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$. Thus by definition this means, $text{inf}_{Nge m} (a_N^{+}) <x$, so $L+varepsilon$ cannot be the infimum of the sequence; and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
(b) Now if $x < L^+=text{inf}_{Nge m} (a_N^{+})le a_N^{+}$. Let us fix some $Nge m$. Thus, $x <a_N^{+}$, and by definition there must be some $nge N$ such that $x< a_n$ (otherwise $x$ would be an upper bound which is less than the least upper bound, which clearly is a contradiction ), as desired. The other part is proven similarly.
(c) Since $text{sup}_{nge m} (a_n) = a_m^+$ and $text{inf}_{nge m} (a_n) = a_m^-$. Clearly, $a_m^-letext{sup}_{Nge m}(a_N^{-})= L^-$ and $L^+=text{inf}_{Nge m} (a_N^{+})le a_m^+$. So, we need to show $L^-le L^+$. Since we have $a_N^{-}le a_N^{+}$ for every $N$. Thus $L^-le L^+$ (because the limit conserve the non-strict inequality), as desired.
(d) Suppose both $L^{+}, L^{-}$ are finite and we let $varepsilon>0$ be given. By (a) there is some $N$ such that $a_n le L^+ +varepsilon$ for each $nge N$. Now, using the definition of limit point, there is some $n_0ge N$ such that $|a_{n_0}-c| le varepsilon$. Thus $,c-L^{+}= (c-a_{n_0})+(a_{n_0}-L^{+})le 2 varepsilon$, since $varepsilon$ was arbitrary, this implies $cle L^{+}$. Similarly, there is some $Mge m$, so that $L^{-}-varepsilon < a_n$ for each $nge M$. By hypothesis $c$ is a limit point and then, for this $M$ there is an $n'ge M$ for which $|a_{n'}-c|le varepsilon$. Thus $L^{-}-c = L^{-} - a_{n'}+a_{n'}-cle 2 varepsilon$, i.e., $L^{-}-cle 2 varepsilon$ for any $varepsilon>0$ and hence $L^{-}le c$.
If $L^{+} = +infty$, then $cle L^{+}$. Now either $L^{-}=+infty$ or $L^{-}=-infty$. The second case is trivial. But in the first case this would imply that the limit point is infinite and we are not defined yet infinite limit points but follows in the exact same way.
(e) Suppose $L^{+}$ is finite and let $varepsilon>0$ be given. Thus, as a consecuence of (a) there is some $N>m$ so that $a_n<L^{+}+varepsilon$ for any $nge N$. Using (b) we know that there at least one $n'ge N$ such that $L^{+}-varepsilon<a_{n'}$. So, $L^{+}-varepsilon< a_{n'}<L^{+}+varepsilon$, i.e., $|a_{n'}-L^{+}|le varepsilon$. Since $varepsilon$ was arbitrary and we can choose any $Nge m$, thus $L^{+}$ is a limit point. The other case is symmetric.
(f) Suppose $(a_n)rightarrow c$; we wish to show $L^{+}= L^{-}$, by (c) we already know that $L^{-}le cle L^{+}$ so, it will suffice to prove $L^{+}le c le L^{-}$.
Let $varepsilon>0$ be given. Since $(a_n)$ converges to a real number, then is bounded. So, the limit superior and inferior are finite and limit points. There is a $Nge m$ such that $|a_n-c|le varepsilon,$ for each $n$, and there is $n'ge N$ so that $L^{+}-varepsilon< a_n'$. Thus $L^{+}-c = (L^{+}-a_n')+(a_n'-c)le 2varepsilon$. And thus $L^{+}le c$. A symmetric argument prove that $L^{-}ge c$ and we're done.
To the other direction we assume that $c=L^{+}=L^{-}$. So, there is some $N_1$ such that $a_n < L^{+} + varepsilon$ and similarly there is some $N_2$ so that $L^{-} - varepsilon< a_n$. We pick the greater between $N_1$ and $N_2$ so, the two inequalities occurs simultaneously. Thus $L^{-} - varepsilon<a_n < L^{+} + varepsilon$ and since both are the same, this would imply $|a_n - c|le varepsilon$ for every $nge N$, which shows that the sequence converges to $c$. $square$
real-analysis proof-verification
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30
add a comment |
Proposition: Let $(a_n)_{n=m}^{infty}$ be a sequence of real numbers, let $L^+$ be the limit superior of the sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+, L^-$ are extened real numbers).
(a) For every $x> L^+$, there exists an $Nge m $ such that $,a_n<x$ for all $nge N$ (the elements of the sequence are eventually less than $x$). Similarly, for every $y<L^-$ there exists an $Nge m $ such that $,a_n>y$ for all $nge N$.
(b) For every $x< L^+$, and every $Nge m$, there exists an $nge N$ such that $a_n> x$ (the elements of the sequence exceed $x$ infinitely often). Similarly for $y>L^-$, and every $Nge m$, there exists an $nge N$ such that $a_n< y$.
(c) We have $,text{inf}_{nge m}(a_n)le L^- le L^+le text{sup}_{nge m}(a_n)$
(d) If $c$ is any limit point of $(a_n)$ then we have $L^- le c le L^+$
(e) If $L^+$ is finite then it is a limit point of the sequence. Similarly if $L^-$ is finite.
(f) Let $c$ be a real number. If $(a_n)rightarrow c$, then we must have $L^+=c=L^-$. Conversely if $L^+=c=L^-$, then $(a_n)rightarrow c$.
I'd like to know if my proof is sound. I'd appreciate any suggestion.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$. Thus by definition this means, $text{inf}_{Nge m} (a_N^{+}) <x$, so $L+varepsilon$ cannot be the infimum of the sequence; and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
(b) Now if $x < L^+=text{inf}_{Nge m} (a_N^{+})le a_N^{+}$. Let us fix some $Nge m$. Thus, $x <a_N^{+}$, and by definition there must be some $nge N$ such that $x< a_n$ (otherwise $x$ would be an upper bound which is less than the least upper bound, which clearly is a contradiction ), as desired. The other part is proven similarly.
(c) Since $text{sup}_{nge m} (a_n) = a_m^+$ and $text{inf}_{nge m} (a_n) = a_m^-$. Clearly, $a_m^-letext{sup}_{Nge m}(a_N^{-})= L^-$ and $L^+=text{inf}_{Nge m} (a_N^{+})le a_m^+$. So, we need to show $L^-le L^+$. Since we have $a_N^{-}le a_N^{+}$ for every $N$. Thus $L^-le L^+$ (because the limit conserve the non-strict inequality), as desired.
(d) Suppose both $L^{+}, L^{-}$ are finite and we let $varepsilon>0$ be given. By (a) there is some $N$ such that $a_n le L^+ +varepsilon$ for each $nge N$. Now, using the definition of limit point, there is some $n_0ge N$ such that $|a_{n_0}-c| le varepsilon$. Thus $,c-L^{+}= (c-a_{n_0})+(a_{n_0}-L^{+})le 2 varepsilon$, since $varepsilon$ was arbitrary, this implies $cle L^{+}$. Similarly, there is some $Mge m$, so that $L^{-}-varepsilon < a_n$ for each $nge M$. By hypothesis $c$ is a limit point and then, for this $M$ there is an $n'ge M$ for which $|a_{n'}-c|le varepsilon$. Thus $L^{-}-c = L^{-} - a_{n'}+a_{n'}-cle 2 varepsilon$, i.e., $L^{-}-cle 2 varepsilon$ for any $varepsilon>0$ and hence $L^{-}le c$.
If $L^{+} = +infty$, then $cle L^{+}$. Now either $L^{-}=+infty$ or $L^{-}=-infty$. The second case is trivial. But in the first case this would imply that the limit point is infinite and we are not defined yet infinite limit points but follows in the exact same way.
(e) Suppose $L^{+}$ is finite and let $varepsilon>0$ be given. Thus, as a consecuence of (a) there is some $N>m$ so that $a_n<L^{+}+varepsilon$ for any $nge N$. Using (b) we know that there at least one $n'ge N$ such that $L^{+}-varepsilon<a_{n'}$. So, $L^{+}-varepsilon< a_{n'}<L^{+}+varepsilon$, i.e., $|a_{n'}-L^{+}|le varepsilon$. Since $varepsilon$ was arbitrary and we can choose any $Nge m$, thus $L^{+}$ is a limit point. The other case is symmetric.
(f) Suppose $(a_n)rightarrow c$; we wish to show $L^{+}= L^{-}$, by (c) we already know that $L^{-}le cle L^{+}$ so, it will suffice to prove $L^{+}le c le L^{-}$.
Let $varepsilon>0$ be given. Since $(a_n)$ converges to a real number, then is bounded. So, the limit superior and inferior are finite and limit points. There is a $Nge m$ such that $|a_n-c|le varepsilon,$ for each $n$, and there is $n'ge N$ so that $L^{+}-varepsilon< a_n'$. Thus $L^{+}-c = (L^{+}-a_n')+(a_n'-c)le 2varepsilon$. And thus $L^{+}le c$. A symmetric argument prove that $L^{-}ge c$ and we're done.
To the other direction we assume that $c=L^{+}=L^{-}$. So, there is some $N_1$ such that $a_n < L^{+} + varepsilon$ and similarly there is some $N_2$ so that $L^{-} - varepsilon< a_n$. We pick the greater between $N_1$ and $N_2$ so, the two inequalities occurs simultaneously. Thus $L^{-} - varepsilon<a_n < L^{+} + varepsilon$ and since both are the same, this would imply $|a_n - c|le varepsilon$ for every $nge N$, which shows that the sequence converges to $c$. $square$
real-analysis proof-verification
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
Proposition: Let $(a_n)_{n=m}^{infty}$ be a sequence of real numbers, let $L^+$ be the limit superior of the sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+, L^-$ are extened real numbers).
(a) For every $x> L^+$, there exists an $Nge m $ such that $,a_n<x$ for all $nge N$ (the elements of the sequence are eventually less than $x$). Similarly, for every $y<L^-$ there exists an $Nge m $ such that $,a_n>y$ for all $nge N$.
(b) For every $x< L^+$, and every $Nge m$, there exists an $nge N$ such that $a_n> x$ (the elements of the sequence exceed $x$ infinitely often). Similarly for $y>L^-$, and every $Nge m$, there exists an $nge N$ such that $a_n< y$.
(c) We have $,text{inf}_{nge m}(a_n)le L^- le L^+le text{sup}_{nge m}(a_n)$
(d) If $c$ is any limit point of $(a_n)$ then we have $L^- le c le L^+$
(e) If $L^+$ is finite then it is a limit point of the sequence. Similarly if $L^-$ is finite.
(f) Let $c$ be a real number. If $(a_n)rightarrow c$, then we must have $L^+=c=L^-$. Conversely if $L^+=c=L^-$, then $(a_n)rightarrow c$.
I'd like to know if my proof is sound. I'd appreciate any suggestion.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$. Thus by definition this means, $text{inf}_{Nge m} (a_N^{+}) <x$, so $L+varepsilon$ cannot be the infimum of the sequence; and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
(b) Now if $x < L^+=text{inf}_{Nge m} (a_N^{+})le a_N^{+}$. Let us fix some $Nge m$. Thus, $x <a_N^{+}$, and by definition there must be some $nge N$ such that $x< a_n$ (otherwise $x$ would be an upper bound which is less than the least upper bound, which clearly is a contradiction ), as desired. The other part is proven similarly.
(c) Since $text{sup}_{nge m} (a_n) = a_m^+$ and $text{inf}_{nge m} (a_n) = a_m^-$. Clearly, $a_m^-letext{sup}_{Nge m}(a_N^{-})= L^-$ and $L^+=text{inf}_{Nge m} (a_N^{+})le a_m^+$. So, we need to show $L^-le L^+$. Since we have $a_N^{-}le a_N^{+}$ for every $N$. Thus $L^-le L^+$ (because the limit conserve the non-strict inequality), as desired.
(d) Suppose both $L^{+}, L^{-}$ are finite and we let $varepsilon>0$ be given. By (a) there is some $N$ such that $a_n le L^+ +varepsilon$ for each $nge N$. Now, using the definition of limit point, there is some $n_0ge N$ such that $|a_{n_0}-c| le varepsilon$. Thus $,c-L^{+}= (c-a_{n_0})+(a_{n_0}-L^{+})le 2 varepsilon$, since $varepsilon$ was arbitrary, this implies $cle L^{+}$. Similarly, there is some $Mge m$, so that $L^{-}-varepsilon < a_n$ for each $nge M$. By hypothesis $c$ is a limit point and then, for this $M$ there is an $n'ge M$ for which $|a_{n'}-c|le varepsilon$. Thus $L^{-}-c = L^{-} - a_{n'}+a_{n'}-cle 2 varepsilon$, i.e., $L^{-}-cle 2 varepsilon$ for any $varepsilon>0$ and hence $L^{-}le c$.
If $L^{+} = +infty$, then $cle L^{+}$. Now either $L^{-}=+infty$ or $L^{-}=-infty$. The second case is trivial. But in the first case this would imply that the limit point is infinite and we are not defined yet infinite limit points but follows in the exact same way.
(e) Suppose $L^{+}$ is finite and let $varepsilon>0$ be given. Thus, as a consecuence of (a) there is some $N>m$ so that $a_n<L^{+}+varepsilon$ for any $nge N$. Using (b) we know that there at least one $n'ge N$ such that $L^{+}-varepsilon<a_{n'}$. So, $L^{+}-varepsilon< a_{n'}<L^{+}+varepsilon$, i.e., $|a_{n'}-L^{+}|le varepsilon$. Since $varepsilon$ was arbitrary and we can choose any $Nge m$, thus $L^{+}$ is a limit point. The other case is symmetric.
(f) Suppose $(a_n)rightarrow c$; we wish to show $L^{+}= L^{-}$, by (c) we already know that $L^{-}le cle L^{+}$ so, it will suffice to prove $L^{+}le c le L^{-}$.
Let $varepsilon>0$ be given. Since $(a_n)$ converges to a real number, then is bounded. So, the limit superior and inferior are finite and limit points. There is a $Nge m$ such that $|a_n-c|le varepsilon,$ for each $n$, and there is $n'ge N$ so that $L^{+}-varepsilon< a_n'$. Thus $L^{+}-c = (L^{+}-a_n')+(a_n'-c)le 2varepsilon$. And thus $L^{+}le c$. A symmetric argument prove that $L^{-}ge c$ and we're done.
To the other direction we assume that $c=L^{+}=L^{-}$. So, there is some $N_1$ such that $a_n < L^{+} + varepsilon$ and similarly there is some $N_2$ so that $L^{-} - varepsilon< a_n$. We pick the greater between $N_1$ and $N_2$ so, the two inequalities occurs simultaneously. Thus $L^{-} - varepsilon<a_n < L^{+} + varepsilon$ and since both are the same, this would imply $|a_n - c|le varepsilon$ for every $nge N$, which shows that the sequence converges to $c$. $square$
real-analysis proof-verification
real-analysis proof-verification
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
edited Sep 5 '17 at 21:39
Jack
27.2k1781198
27.2k1781198
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
asked Nov 5 '13 at 2:05
Jose Antonio
4,40421527
4,40421527
For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30
add a comment |
For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30
For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30
add a comment |
2 Answers
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So many questions. Let's look at your proof for the first one.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$.
Thusby definition(of$L^+$)this(OK, "this" refers to$x>L^+$)means,
$$
text{inf}_{Nge m} (a_N^{+}) <x,
$$
(This is a wrong expression. The left hand side of the inequality should not depend on$m$. You should have instead$inf_{N} (a_N^{+}) <x$.)
so $L+varepsilon$ cannot be the infimum of the sequence;
(This is very confusing: (1) typo in$L+varepsilon$: a superscript is missing; (2) what does "the sequence" refer to? (3) Assuming the sequence refers to$(a_N^+)$, the fact that$L^+varepsilon$cannot be its infimum follows directly from the definition of$L^+$and I don't understand why you need the above argument to say "so". I would stop reading from here.)
and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
answered Sep 5 '17 at 21:55
Jack
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I don't think the proof of (e) ["if $L^+$ is finite then it is a limit point of the sequence"] was entirely correct, since the choice of $N$ wasn't actually arbitrary. The original proof didn't use the fact that $L^+$ is finite either, which may suggest problems.
Improved version:
Since $L^+$ is finite, it can be shown that $L^+=lim_{ntoinfty} (a^+_n)_{n=m}^infty =inf_{ngeq m} a^+_n$. Now fix any $epsilon >0$ and $N in mathbb{N}, Ngeq m$. The subsequence $(a^+_n)_{n=N}^infty$ of $ (a^+_n)_{n=m}^infty$ must also converge to $L^+$, which implies $(a^n)_{n=N}^infty$ also has $L^+$ as its limsup. Now applying (a) on $(a^n)_{n=N}^infty$ gives the existence of some $N_1 geq N$ s.t. $forall ngeq N_1, a_n <L^+ + epsilon$; then (b) gives the existence of some $n_2 geq N_1$ s.t. $L^+ - epsilon < a_{n_2}$. So $L^+ - epsilon < a_{n_2} <L^+ + epsilon $, we've found some $n_2 geq N$ s.t. $|a_{n_2} - L^+ | leq epsilon$, for arbitrary $epsilon >0$ and $Ngeq m$.
answered Jul 15 '18 at 3:48
Yibo Yang
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So many questions. Let's look at your proof for the first one.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$.
Thusby definition(of$L^+$)this(OK, "this" refers to$x>L^+$)means,
$$
text{inf}_{Nge m} (a_N^{+}) <x,
$$
(This is a wrong expression. The left hand side of the inequality should not depend on$m$. You should have instead$inf_{N} (a_N^{+}) <x$.)
so $L+varepsilon$ cannot be the infimum of the sequence;
(This is very confusing: (1) typo in$L+varepsilon$: a superscript is missing; (2) what does "the sequence" refer to? (3) Assuming the sequence refers to$(a_N^+)$, the fact that$L^+varepsilon$cannot be its infimum follows directly from the definition of$L^+$and I don't understand why you need the above argument to say "so". I would stop reading from here.)
and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
add a comment |
So many questions. Let's look at your proof for the first one.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$.
Thusby definition(of$L^+$)this(OK, "this" refers to$x>L^+$)means,
$$
text{inf}_{Nge m} (a_N^{+}) <x,
$$
(This is a wrong expression. The left hand side of the inequality should not depend on$m$. You should have instead$inf_{N} (a_N^{+}) <x$.)
so $L+varepsilon$ cannot be the infimum of the sequence;
(This is very confusing: (1) typo in$L+varepsilon$: a superscript is missing; (2) what does "the sequence" refer to? (3) Assuming the sequence refers to$(a_N^+)$, the fact that$L^+varepsilon$cannot be its infimum follows directly from the definition of$L^+$and I don't understand why you need the above argument to say "so". I would stop reading from here.)
and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
add a comment |
So many questions. Let's look at your proof for the first one.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$.
Thusby definition(of$L^+$)this(OK, "this" refers to$x>L^+$)means,
$$
text{inf}_{Nge m} (a_N^{+}) <x,
$$
(This is a wrong expression. The left hand side of the inequality should not depend on$m$. You should have instead$inf_{N} (a_N^{+}) <x$.)
so $L+varepsilon$ cannot be the infimum of the sequence;
(This is very confusing: (1) typo in$L+varepsilon$: a superscript is missing; (2) what does "the sequence" refer to? (3) Assuming the sequence refers to$(a_N^+)$, the fact that$L^+varepsilon$cannot be its infimum follows directly from the definition of$L^+$and I don't understand why you need the above argument to say "so". I would stop reading from here.)
and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
So many questions. Let's look at your proof for the first one.
Proof: Let $a_N^{+}=text{sup}_{nge N} (a_n)$ and $a_N^{-}=text{inf}_{nge N} (a_n)$.
(a) Suppose $x> L^+$. Let $varepsilon = x-L^+$.
Thusby definition(of$L^+$)this(OK, "this" refers to$x>L^+$)means,
$$
text{inf}_{Nge m} (a_N^{+}) <x,
$$
(This is a wrong expression. The left hand side of the inequality should not depend on$m$. You should have instead$inf_{N} (a_N^{+}) <x$.)
so $L+varepsilon$ cannot be the infimum of the sequence;
(This is very confusing: (1) typo in$L+varepsilon$: a superscript is missing; (2) what does "the sequence" refer to? (3) Assuming the sequence refers to$(a_N^+)$, the fact that$L^+varepsilon$cannot be its infimum follows directly from the definition of$L^+$and I don't understand why you need the above argument to say "so". I would stop reading from here.)
and thus there is at least one $Nge K$ ($Kge m$) such that $a_N^{+}<L+varepsilon = x$. Since $,a_n le a_N^{+}$. Thus $a_n le x $ for each $nge N$. The second part can be proved using a similar argument.
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
edited Sep 5 '17 at 22:02
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
answered Sep 5 '17 at 21:55
Jack
27.2k1781198
27.2k1781198
add a comment |
add a comment |
I don't think the proof of (e) ["if $L^+$ is finite then it is a limit point of the sequence"] was entirely correct, since the choice of $N$ wasn't actually arbitrary. The original proof didn't use the fact that $L^+$ is finite either, which may suggest problems.
Improved version:
Since $L^+$ is finite, it can be shown that $L^+=lim_{ntoinfty} (a^+_n)_{n=m}^infty =inf_{ngeq m} a^+_n$. Now fix any $epsilon >0$ and $N in mathbb{N}, Ngeq m$. The subsequence $(a^+_n)_{n=N}^infty$ of $ (a^+_n)_{n=m}^infty$ must also converge to $L^+$, which implies $(a^n)_{n=N}^infty$ also has $L^+$ as its limsup. Now applying (a) on $(a^n)_{n=N}^infty$ gives the existence of some $N_1 geq N$ s.t. $forall ngeq N_1, a_n <L^+ + epsilon$; then (b) gives the existence of some $n_2 geq N_1$ s.t. $L^+ - epsilon < a_{n_2}$. So $L^+ - epsilon < a_{n_2} <L^+ + epsilon $, we've found some $n_2 geq N$ s.t. $|a_{n_2} - L^+ | leq epsilon$, for arbitrary $epsilon >0$ and $Ngeq m$.
answered Jul 15 '18 at 3:48
Yibo Yang
505316
add a comment |
I don't think the proof of (e) ["if $L^+$ is finite then it is a limit point of the sequence"] was entirely correct, since the choice of $N$ wasn't actually arbitrary. The original proof didn't use the fact that $L^+$ is finite either, which may suggest problems.
Improved version:
Since $L^+$ is finite, it can be shown that $L^+=lim_{ntoinfty} (a^+_n)_{n=m}^infty =inf_{ngeq m} a^+_n$. Now fix any $epsilon >0$ and $N in mathbb{N}, Ngeq m$. The subsequence $(a^+_n)_{n=N}^infty$ of $ (a^+_n)_{n=m}^infty$ must also converge to $L^+$, which implies $(a^n)_{n=N}^infty$ also has $L^+$ as its limsup. Now applying (a) on $(a^n)_{n=N}^infty$ gives the existence of some $N_1 geq N$ s.t. $forall ngeq N_1, a_n <L^+ + epsilon$; then (b) gives the existence of some $n_2 geq N_1$ s.t. $L^+ - epsilon < a_{n_2}$. So $L^+ - epsilon < a_{n_2} <L^+ + epsilon $, we've found some $n_2 geq N$ s.t. $|a_{n_2} - L^+ | leq epsilon$, for arbitrary $epsilon >0$ and $Ngeq m$.
answered Jul 15 '18 at 3:48
Yibo Yang
505316
add a comment |
I don't think the proof of (e) ["if $L^+$ is finite then it is a limit point of the sequence"] was entirely correct, since the choice of $N$ wasn't actually arbitrary. The original proof didn't use the fact that $L^+$ is finite either, which may suggest problems.
Improved version:
Since $L^+$ is finite, it can be shown that $L^+=lim_{ntoinfty} (a^+_n)_{n=m}^infty =inf_{ngeq m} a^+_n$. Now fix any $epsilon >0$ and $N in mathbb{N}, Ngeq m$. The subsequence $(a^+_n)_{n=N}^infty$ of $ (a^+_n)_{n=m}^infty$ must also converge to $L^+$, which implies $(a^n)_{n=N}^infty$ also has $L^+$ as its limsup. Now applying (a) on $(a^n)_{n=N}^infty$ gives the existence of some $N_1 geq N$ s.t. $forall ngeq N_1, a_n <L^+ + epsilon$; then (b) gives the existence of some $n_2 geq N_1$ s.t. $L^+ - epsilon < a_{n_2}$. So $L^+ - epsilon < a_{n_2} <L^+ + epsilon $, we've found some $n_2 geq N$ s.t. $|a_{n_2} - L^+ | leq epsilon$, for arbitrary $epsilon >0$ and $Ngeq m$.
answered Jul 15 '18 at 3:48
Yibo Yang
505316
I don't think the proof of (e) ["if $L^+$ is finite then it is a limit point of the sequence"] was entirely correct, since the choice of $N$ wasn't actually arbitrary. The original proof didn't use the fact that $L^+$ is finite either, which may suggest problems.
Improved version:
Since $L^+$ is finite, it can be shown that $L^+=lim_{ntoinfty} (a^+_n)_{n=m}^infty =inf_{ngeq m} a^+_n$. Now fix any $epsilon >0$ and $N in mathbb{N}, Ngeq m$. The subsequence $(a^+_n)_{n=N}^infty$ of $ (a^+_n)_{n=m}^infty$ must also converge to $L^+$, which implies $(a^n)_{n=N}^infty$ also has $L^+$ as its limsup. Now applying (a) on $(a^n)_{n=N}^infty$ gives the existence of some $N_1 geq N$ s.t. $forall ngeq N_1, a_n <L^+ + epsilon$; then (b) gives the existence of some $n_2 geq N_1$ s.t. $L^+ - epsilon < a_{n_2}$. So $L^+ - epsilon < a_{n_2} <L^+ + epsilon $, we've found some $n_2 geq N$ s.t. $|a_{n_2} - L^+ | leq epsilon$, for arbitrary $epsilon >0$ and $Ngeq m$.
answered Jul 15 '18 at 3:48
Yibo Yang
505316
answered Jul 15 '18 at 3:48
Yibo Yang
505316
answered Jul 15 '18 at 3:48
Yibo Yang
505316
answered Jul 15 '18 at 3:48
Yibo Yang
505316
505316
add a comment |
add a comment |
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For (d), in your definition of limit point, it should be clearer that such an $epsilon$ exists for each $n_0 ge N$.
– user203509
Apr 29 '16 at 22:03
I think the proof of (d) is correct; it was shown that for any arbitrary $epsilon>0, c-L^+leq 2 epsilon$, thus $c-L^+leq 0$.
– Yibo Yang
Jul 15 '18 at 3:30