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The proposition comes from the textbook of Tao Analysis:

Proposition 6.4.12. Let $(a_n)_{n=m}^∞ $ be a sequence of real numbers, let $L^+$ be the limit superior of this sequence, and let $L^-$ be the limit inferior of this sequence (thus both $L^+$ and $L^-$ are extended real numbers).
Let $c$ be a real number. If $(a_n)_{n=m}^∞ $ converges to $c$, then we must have $L^+$ = $L^-$ = $c$. Conversely, if $L^+$ = $L^-$ = $c$, then $(a_n)_{n=m}^∞ $ converges to $c$.

My question is: Can we set c to be an extended real number? So if $L^+$ = $L^-$ = $c$=$+\infty$, then $(a_n)_{n=m}^∞ $ converges to $+\infty$. I'm just wondering if the limit of a sequence can be $+\infty$. If yes, then the sequence (1,2,3,$\dots$) has a limit of $+\infty$.

Angelo
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Andrew Li
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  • Are you essentially asking why Tao only considers real $c$ as opposed to extended real $c$? – Brian Moehring Jan 21 '23 at 03:37
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    You cannot take $c=+\infty$ since $c$ a point in $\mathbb{R}$ and not in $\Bbb{R}\cup{\pm\infty}.$ – Nik Jan 21 '23 at 03:40
  • yes, can we set c to an extended real number? so we have limit of sequence (1,2,3,...) be $+\infty$, like in the supremum of a set or sequence can be $+\infty$ @BrianMoehring – Andrew Li Jan 21 '23 at 03:46
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    @AndrewLi, that proposition is correct even if $c=+\infty$ or $c=-\infty$ and in these cases we say that the sequence diverges to infinity. – Angelo Jan 21 '23 at 04:08

1 Answers1

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Answering your questions in reverse order:

I'm just wondering if the limit of a sequence can be $+\infty$. If yes, then the sequence $(1,2,3,\ldots)$ has a limit of $+\infty$.

This is perfectly fine. It's a true statement that $(1,2,3,\ldots)$ has a limit of $+\infty$ in the extended real numbers.

Can we set $c$ to be [any] extended real number? So if $L^+ = L^- = c=+\infty$, then $(a_n)_{n=m}^\infty$ converges to $+\infty$.

This, however, is not okay. By convention, if a sequence of real numbers converges, then its limit is a real number. Therefore, the phrase "$(a_n)_{n=m}^\infty$ converges to $+\infty$" is either false or nonsense. In the former case, your statement following "So" is false when $L^-=c=+\infty$, and in the latter case, it's not even a logical statement.

You may write the following analogous statement for the infinite extended real case:

Let $c \in \{-\infty, +\infty\}$ be an extended real number. If $(a_n)_{n=m}^\infty$ diverges to $c$, then we must have $L^+ = L^- = c$. Conversely, if $L^+ = L^- = c$, then $(a_n)_{n=m}^\infty$ diverges to $c$.

Brian Moehring
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