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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!

Henno Brandsma
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LMW
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    Let me also add to Henno Brandsma's answer: A convergent net $(x_i)$ in a normed space is eventually norm-bounded, i.e. there is $i_0$ such that $\sup_{i \geq i_0} \lVert x_i \rVert < \infty$. But if you only have a weakly convergent net, then it need not be eventually weakly-bounded: https://math.stackexchange.com/questions/158902/must-a-weakly-or-weak-convergent-net-be-eventually-bounded – yada Feb 17 '20 at 08:25

1 Answers1

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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.

Henno Brandsma
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