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 $N\ge m $ such that $\,a_n<x$ for all $n\ge N$ (the elements of the sequence are eventually less than $x$). Similarly, for every $y<L^-$ there exists an $N\ge m $ such that $\,a_n>y$ for all $n\ge N$.

(b) For every $x< L^+$, and every $N\ge m$, there exists an $n\ge N$ such that $a_n> x$ (the elements of the sequence exceed $x$ infinitely often). Similarly for $y>L^-$, and every $N\ge m$, there exists an $n\ge N$ such that $a_n< y$.

(c) We have $\,\text{inf}_{n\ge m}(a_n)\le L^- \le L^+\le \text{sup}_{n\ge m}(a_n)$

(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 $a_N^{+}=\text{sup}_{n\ge N} (a_n)$ and $a_N^{-}=\text{inf}_{n\ge N} (a_n)$.

(a) Suppose $x> L^+$. Let $\varepsilon = x-L^+$. Thus by definition this means, $\text{inf}_{N\ge m} (a_N^{+}) <x$, so $L+\varepsilon$ cannot be the infimum of the sequence; and thus there is at least one $N\ge K$ ($K\ge m$) such that $a_N^{+}<L+\varepsilon = x$. Since $\,a_n \le a_N^{+}$. Thus $a_n \le x $ for each $n\ge N$. The second part can be proved using a similar argument.

(b) Now if $x < L^+=\text{inf}_{N\ge m} (a_N^{+})\le a_N^{+}$. Let us fix some $N\ge m$. Thus, $x <a_N^{+}$, and by definition there must be some $n\ge N$ 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 $\text{sup}_{n\ge m} (a_n) = a_m^+$ and $\text{inf}_{n\ge m} (a_n) = a_m^-$. Clearly, $a_m^-\le\text{sup}_{N\ge m}(a_N^{-})= L^-$ and $L^+=\text{inf}_{N\ge m} (a_N^{+})\le a_m^+$. 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 $n\ge N$. Now, using the definition of limit point, there is some $n_0\ge N$ 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 $c\le L^{+}$. Similarly, there is some $M\ge m$, so that $L^{-}-\varepsilon < a_n$ for each $n\ge M$. 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 $L^{-}-c = L^{-} - a_{n'}+a_{n'}-c\le 2 \varepsilon$, i.e., $L^{-}-c\le 2 \varepsilon$ for any $\varepsilon>0$ and hence $L^{-}\le c$.

If $L^{+} = +\infty$, then $c\le L^{+}$. 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 $n\ge N$. 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 $N\ge m$, 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 $L^{-}\le c\le L^{+}$ 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 $N\ge m$ such that $|a_n-c|\le \varepsilon\,$ for each $n$, and there is $n'\ge N$ so that $L^{+}-\varepsilon< a_n'$. Thus $L^{+}-c = (L^{+}-a_n')+(a_n'-c)\le 2\varepsilon$. 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 $n\ge N$, which shows that the sequence converges to $c$. $\square$

  • $\begingroup$ For (d), in your definition of limit point, it should be clearer that such an $\epsilon$ exists for each $n_0 \ge N$. $\endgroup$ – user203509 Apr 29 '16 at 22:03
  • 1
    $\begingroup$ I think the proof of (d) is correct; it was shown that for any arbitrary $\epsilon>0, c-L^+\leq 2 \epsilon$, thus $c-L^+\leq 0$. $\endgroup$ – Yibo Yang Jul 15 '18 at 3:30
  • $\begingroup$ I would appreciate it if you mentioned the book you got these theorems and questions from, as it sends respect to the author. I'm pretty sure these are from Terence Tao's book on Analysis I as I myself have read it. Please do add that. $\endgroup$ – Pratik Apshinge Aug 7 at 19:03

So many questions. Let's look at your proof for the first one.

Proof: Let $a_N^{+}=\text{sup}_{n\ge N} (a_n)$ and $a_N^{-}=\text{inf}_{n\ge N} (a_n)$.

(a) Suppose $x> L^+$. Let $\varepsilon = x-L^+$. Thus by definition (of $L^+$) this (OK, "this" refers to $x>L^+$) means, $$ \text{inf}_{N\ge m} (a_N^{+}) <x, $$ (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 $N\ge K$ ($K\ge m$) such that $a_N^{+}<L+\varepsilon = x$. Since $\,a_n \le a_N^{+}$. Thus $a_n \le x $ for each $n\ge N$. The second part can be proved using a similar argument.


For the proof of $(c)$ ($L^{-}\leq L^{+}$) is correct only if it is known that $L^{+}$ and $L^{-}$ are finite. So, it would be better to prove the result by not assuming that they are limits of certain sequences (only true if they are finite) but using only the properties of supremum and infimum (of extended real numbers).

$\textbf{Claim:}$ Let $(a_{n})_{n=m}^{\infty}$ and $(b_{n})_{n=m}^{\infty}$ are sequences of extended real numbers. Let $(a_{n})_{n=m}^{\infty}$ be a monotonically increasing sequence and $(b_{n})_{n=m}^{\infty}$ be a monotonically decreasing sequence and $\forall n\geq m$ we have $b_{n}\geq a_{n}$. Then $\inf(b_{n})_{n=m}^{\infty}\geq\sup(a_{n})_{n=m}^{\infty}$.

$\textbf{Proof:}$ Let us assume for the sake of contradiction that $\inf(b_{n})_{n=m}^{\infty} < \sup(a_{n})_{n=m}^{\infty}$ $$\implies\exists n\geq m\:\text{s.t.}\:\inf(b_{n})_{n=m}^{\infty} < a_{n}$$ $$\implies\exists n'\geq m\:\text{s.t.}\:b_{n'} < a_{n}$$ By hypothesis, we have $a_{n}\leq b_{n}\implies b_{n'} < b_{n}\implies n'>n$ since $(b_{n})_{n=m}^{\infty}$ is decreasing. Similarly, $a_{n'}\leq b_{n'}\implies a_{n'} < a_{n}\implies n'<n$ since $(a_{n})_{n=m}^{\infty}$ is increasing. Since both these conditions cannot aoccur simultaneously we have reached a contradiction.

The result may now be proved by applying the claim to the sequences $(a_{n}^{+})_{n=m}^{\infty}$ and $(a_{n}^{-})_{n=m}^{\infty}$.


Your proof for (d) is correct, no arguing that, but I must say there is much simpler way of doing it:

Let $c$ be a limit point for the sequence $a_n$ and suppose $L^->c$. Then, $c<L^-≤L^+$ (proved in (c)).

Since $c<L^-$, there exists an $N≥m$ s.t. $\forall n≥N, a_n > c$.(by (a))

Since $c<L^+$ as well, for every $N'≥m$, there exists $n'≥N' s.t. a_{n'}<c$.(by (b))

But, when $N=N'$, a contradiction occurs. So, $L^-≤c$. Similar and symmetric proof for $c≤L^+$.


Here's my proof of (d). If anyone working through Tao's book happens to see this, please comment if it seems clearer to you as it does to me. I'll prove just one-half of it: $L^-\leq c$. Note that if $c$ is a limit point then by definition it is a real number, and not $\pm\infty$.

Proof: If $L^-=-\infty$, the result is trivial. So suppose $L^{-}$ is a real number or $L^-=\infty$. Assume for contradiction that $c<L^-$. Let $d$ be a real number such that $c<d<L^-$ (if $L^-=\infty$ just let $d=c+1$).

We apply part (a) to $d$. So there exists $N\geq m$ such that $a_n>d>c$ for all $n\geq N$.

Let $\epsilon=\frac{d-c}{2}>0$. Then since $c$ is a limit point, there exists $n\geq N$ such that $a_n-d+2\epsilon=a_n-d+(d-c)=a_n-c=\lvert a_n-c \rvert\leq\epsilon$. But then $a_n-d\leq -\epsilon<0$, a contradiction. Therefore $c\geq L^-$.


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 $L^+=\lim_{n\to\infty} (a^+_n)_{n=m}^\infty =\inf_{n\geq m} a^+_n$. Now fix any $\epsilon >0$ and $N \in \mathbb{N}, N\geq m$. 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. $\forall n\geq N_1, a_n <L^+ + \epsilon$; 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 $N\geq m$.

  • $\begingroup$ I know this might be years later, but, it seems like what the asker did to answer (e) was right. Your right to say the original proof didn't use the fact that $L^+$ is finite; it didn't need to. It says that for every $x$ greater than $L^+$, so the further proof would not apply when $L^+$ is infinite; what I am trying to say is that it didn't need to explicitally say that $L^+$ was finite as it was already implied. $\endgroup$ – Pratik Apshinge Aug 7 at 18:50
  • $\begingroup$ Next, you say that the choice of $N$ wasn't completely arbitrary, which is false. First, the proof says that there exists a $N$, then it says for all $N'$, according to the conditions given above. Since you know this $N$ exists, take all the $N^*≥N$, where $N^*$ is a natural number. For all $n^*≥N^*$, $L^+ - \epsilon ≤ a_{n^*}≤L^+ + \epsilon$. Clearly, there is nothing wrong in this statement, and you have proved your proof. Arbitrariness of N comes from the fact that you've chosen one such one by knwing it existing, but you don't know what value it has. $\endgroup$ – Pratik Apshinge Aug 7 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.