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I've been asked to show that $$ \limsup_{n\ge 1}x_n = \inf\{y\in\mathbb{R}\ |\ \exists N>0\ \text{s.t.}\ x_n< y\ \forall n\ge N\} $$ for bounded sequences $x_n$.

Last night I came up with a rather uninspired argument to prove this that just hinged on showing that the LHS above is the greatest lower bound of the set on the right. I don't think it's a very interesting or clever proof, and I haven't even checked it in a clear state of mind so it might not be completely correct, but I was wondering if someone could verify if I'm in the right direction.

Additionally, if the argument is correct, but overcomplicated or contrived, I would appreciate a hint pointing to a nicer solution!

Proof

Let $z_n = \sup_{k\ge n}x_n$, and $\Gamma = \{\gamma\in\mathbb{R}\ |\ \exists N>0\ \text{s.t.}\ \forall n\ge N,\ x_n<\gamma\}$. We will show that for all $\gamma\in\Gamma$, $\lim_{n\to\infty}z_n \le \gamma$, and that if $z\in\mathbb{R}$ is such that $z\le \gamma$ for all $\gamma\in\Gamma$, then $z\le\lim_{n\to\infty}z_n$.

Take $\gamma\in\Gamma$. To show that $\lim_{n\to\infty}z_n\le \gamma$, we need to show that $\exists N\ \text{s.t.}\ \forall n\ge N,\ z_n\le\gamma$. However, $\gamma\in\Gamma$ implies $\exists N = N_0\ \text{s.t.}\ \forall n\ge N,\ x_n\le \gamma$. Then for all $n\ge N_0$, $x_n\le\gamma$, implying $\sup_{k\ge n}x_n \le \gamma$, and so $z_n\le\gamma$ for all $n\ge N_0$. Thus, $\lim_{n\to\infty} z_n\le \gamma$, so $\limsup_{n\ge 1}x_n$ is a lower bound for $\Gamma$.

Now, let $z\in\mathbb{R}\backslash\Gamma$ be an arbitrary lower bound for $\Gamma$. Suppose $\limsup_{n\ge 1}x_n < z$. Then $\exists N$ such that for all $n\ge N$, $\sup_{k\ge n}x_n < z$. For this same $N$, for all $n\ge N$, $x_n \le \sup_{k\ge n}x_n < z$. This, however, implies $z\in\Gamma$, a contradiction. Thus $\limsup_{n\ge 1}x_n \ge z$ for all lower bounds $z\not\in\Gamma$.

If $\gamma = \inf(\Gamma)\in\Gamma$, then $\exists z_n\in B_{1/n}(\gamma)\backslash\Gamma$ for all $n\ge N$ which is a lower bound for $\Gamma$ not in $\Gamma$, and $\limsup_{n\ge 1}x_n\ge z_n$ for all $n$. Since this is true for arbitrary $n$, we can only conclude that $\limsup_{n\ge 1}x_n = \gamma$.

user3002473
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  • The right-hand-side may not exist if the $\limsup$ is infinity. – Michael May 27 '19 at 14:31
  • @Michael Sorry I should mention that the sequence $x_n$ is bounded, and so by MCT $\lim_{n\to\infty}\sup_{k\ge n}x_n$ converges. – user3002473 May 27 '19 at 14:37
  • Is your last paragraph supposed to be explaining why accomplishing the goals of the first paragraph is good? – Michael May 27 '19 at 15:01
  • Your proof is likely correct (I'm not sure about the last paragraph). You might make it conceptually easier by breaking it into (i) Show $\limsup \leq \inf$; (ii) Show $\limsup \geq \inf$. I think your paragraphs 2 and 3 have the essential ingredients for this if the reader interprets them as such. – Michael May 27 '19 at 15:04
  • @Michael well, my second last paragraph shows that for any lower bound of $\Gamma$, which is not contained in $\Gamma$, $\limsup_{n\ge 1}x_n$ is greater than or equal to this lower bound. So the last paragraph handles the case where the lower bound is in $\Gamma$. It's a little redundant, so I'll fiddle with your idea of showing $\limsup \le $ and $\ge \inf$. – user3002473 May 27 '19 at 15:35
  • @Michael, I think it makes sense to interpret that inf of the null set as $\infty$. So the equality holds true even when is $\infty$. – Julian Mejia May 27 '19 at 15:37
  • @JulianMejia If the infimum of the empty set is $\infty$, what is the supremum? If $\sup = -\infty$ then $\sup < \inf$, which is a bit odd. – Michael May 27 '19 at 15:55
  • @Michael Indeed a bit odd, but nevertheless commonly accepted. Have a look at this question and its answers. – drhab May 28 '19 at 09:54

2 Answers2

1

Some minor structural suggestions that may simplify things.

[Your posted proof has 4 paragraphs, I refer to the second and third.]

1) Preliminary:

Note that boundedness of $\{x_n\}_{n=1}^{\infty}$ implies the set $\Gamma$ is non-empty and lower-bounded. Define $\gamma^*=\inf \Gamma$ and note that $\gamma^*$ is finite.

2) Show $\limsup_{n\rightarrow\infty} x_n\leq \gamma^*$:

Your paragraph 2 already shows $$ \limsup_{n\rightarrow\infty} x_n \leq \gamma \quad \forall \gamma \in \Gamma$$ Taking the "infimum of both sides" concludes $$ \limsup_{n\rightarrow\infty} x_n \leq \gamma^* $$

3) Show $\limsup_{n\rightarrow\infty} x_n \geq \gamma^*$:

Suppose not (we reach a contradiction). Then the $\limsup$ is strictly less than $\gamma^*$ and there is a number $z$ in between: $$\limsup_{n\rightarrow\infty} x_n < z< \gamma^*$$ Your paragraph 3 already shows this yields a contradiction.

Michael
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Defining $s_{n}=\sup_{k\geq n}x_{k}$ we have $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$ so it is direct that: $$\limsup x_{n}=\lim_{n\to\infty}s_{n}=\inf\left\{ s_{n}\mid n\in\mathbb{N}\right\} $$

It remains to be shown that the sets

  • $A:=\left\{ s_{n}\mid n\in\mathbb{N}\right\}$
  • $B:=\left\{ y\in\mathbb{R}\mid\exists N\in\mathbb{N}\forall n[n>N\implies x_{n}<y\right\}$

have the same infimum or equivalently that the set of lower bounds of $A$ coincides with the set of lower bounds of $B$.

If $z$ is not a lower bound of $A$ then some $n$ exists with $s_{n}<z$.

Then we can choose some $y$ satisfying $s_{n}<y<z$.

We have $x_{k}<y$ for every $k\geq n$ showing that $y\in B\wedge y<z$ and conclude that $z$ is not a lower bound of $B$.

If $z$ is not a lower bound of $B$ then some $y\in B$ exists with $y<z$ and from $y\in B$ we conclude that some $n$ exists with $s_{n}\leq y$.

Then $s_{n}<z$ and we conclude that $z$ is not a lower bound of $A$.

This completes the proof.

drhab
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