If $ X $ is a normed space and $ (x_n)_{n=1}^{\infty} \subset X $ is a convergent sequence, then it is elementary to show that $ \| x_n \| $ is bounded by observing that there exists an $ N \in \mathbb{N} $ such that $ \| x_n – x \| \leq 1 $ for all $ n \geq N $, and so:
$$ \| x_n \| \leq \max \{ \| x_1 \|, \| x_2 \|, …, \| x_{N-1} \|, \| x \| + 1 \} $$
But if $ (x_i)_{i \in I} \subset X $ is a net that converges to $ x \in X $, can we conclude that $ (x_i)_{i \in I} $ or $ (x_i – x)_{i \in I} $ is bounded? If not, is there any additional conditions on $ X $ or the net $ (x_i)_{i \in I} $ which will guarantee convergence implies boundedness?
Thanks!
Best Answer
There is no way to guarantee that the whole net is bounded: a sequence has the special property that it is a net on an index set such that each initial segment is finite. This allows your argument to work. But for a convergent net, like for a convergent sequence, we can only control what happens in a tail...
Simple example: take $I= (\mathbb{Z}, \le)$ which is a directed set (even a linear order). Define the following net in $\mathbb{R}$: $x_i = i$ if $i <0$, and $x_i=0$ for $i\ge 0$. Then this net converges to $0$ (we can take $i_0=0$ for any neighbourhood of $0$) but the image of the net is $\{0,-1,-2,-3,-4,\ldots\}$, which is unbounded, because we cannot "control" initial segments.