Every convergent sequence is bounded: what's wrong with this counterexample? 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!
 A: 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.
A: 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: 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 $1/0=\infty$.
A: $x_n = 1 / (n - a)$ for some natural number n is not a sequence of real numbers as all, since $x_a$ is not defined. 
If you picked for example $a = 1000 - 10^{-1000}$, then $x_{1000} = 10^{1000}$. The sequence is convergent to 0, and while $x_{1000}$ is pretty large, it is still bounded by $10^{1000}$. 
The simple idea behind the theorem is that the sequence will have an infinite number of elements that are not much larger than the limit, and only a finite number of elements that are not quite close to the limit. The numbers close to the limit are bounded because they are not much larger than the limit, the numbers further away from the limit are bounded because there is only a finite number of them. 
There's no cheating by introducing a number to the sequence that would be unbounded on its own - that's not a real number. 
A: I find this absurdly circular. If $x_n = 1/(n−a)$ is undefined at $n=a$, for $n$, $a$ $\in \mathbb{N}$, then the sequence is defined to not be infinite — i.e. finite and bounded — from the get go. Why then do we need a proof that $x_n$ is bounded because it converges? Any sequence is bounded given this definition.
For example, if we take $y_n = 1/x_n = (n−a)$ then, given the arguments above, $\pm \infty$ will be omitted from the range of $y_n$. Thus, $y_n$ is finite and therefore bounded.
