A basic result in analysis states that convergence of a sequence implies its boundedness. I was wondering: what's wrong with $x_n = 1/(n-a)$ for some $a \in N$? This sequence is convergent to $0$, but $x_a$ is unbounded. What am I missing here? Thanks!
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7How do you define $x_{a}$? If you plug $n=a$ you get $x_{a}=1/0$. – Prism 6 hours ago
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2$x_a$ is undefined so that is not a sequence at all. – fleablood 6 hours ago
The result is saying that any convergence sequence in real numbers is bounded. The sequence that you have constructed is not a sequence in real numbers, it is a sequence in extended real numbers if you take the convention that $1/0=\infty$.
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Okay, I see. So a sequence of real numbers can be unbounded only asymptotically? – Tobia Marcucci 6 hours ago
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I mean, if you really want to say a sequence of real numbers, then you have to make sure all terms are defined. But if you take the convention that $1/0=\infty$, it is not a sequence of real numbers because $\infty$ is not in the system of real numbers. – user284331 6 hours ago
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I mean that, if you rule out all the cases like the one I asked about, an element of a sequence can be unbounded only as $n$ goes to infinity (e.g. $x_n=n$), and it's never the case that a sequence is unbounded because one of its elements $x_n$ is unbounded with $n$ finite. – Tobia Marcucci 6 hours ago
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1Of course, for a finite set of real numbers must be bounded, just take their maxima. – user284331 6 hours ago
Your sequence $\{ 1/(n-a)\}_1^\infty $ is not defined at $n=a$ if $a$ is a positive integer.
Thus it is not a sequence at all.
for example $$\{ 1/(n-5)\} _1^\infty = \{ -1/4,-1/3,-1/2, -1, ?, 1,...\}$$
A real sequence is nothing but a function $$f:\mathbb{N}\longrightarrow \mathbb{R}$$ One often writes for example $x_n$ instead of $f(n)$ etc.
So, your "sequence" is not defined on $\mathbb{N}$ but on $\mathbb{N}\setminus\{a\}$.
But nicely enough, if you remove the "undefined member" by setting $x_a := r$ to an arbitrary real number $r$, you get a convergent sequence which is bounded.
