There have been several works characterizing weak topology by nonstandard analysis, which give rise to the following monad ($X$ is a Hilbert space):
$$\mu(0) = \{y\in{}^{\*}X: \forall x\in X ~~ \langle y|x\rangle\simeq0\}.$$
That is collection of vectors (possibly non standard and not near-standard) that have infinitesimal inner product with all other standard vectors.


However, what can we say about norm of vectors in $\mu(0)$?
Is the following statement true?
$$\forall y \in \mu(0), \exists r \in \mathbb{R}, |y|<r$$
This means every vector in weak monad has finite norm.


This statement implies uniform boundedness theorem. Uniform boundedness theorem means every set that is weakly bounded is bounded. In nonstandard analysis characterization, that means every standard set that can be absorbed by a weak monad can also be absorbed by ordinary monad. The above statement just says a weak monad is contained in ordinary monad times any infinite number.


If this statement is wrong, can we construct any counter example?




---


*Anderson, Robert M.; Rashid, Salim*, [**A nonstandard characterization of weak convergence**](http://dx.doi.org/10.2307/2042621), Proc. Am. Math. Soc. 69, 327-332 (1978). [ZBL0393.03047](https://zbmath.org/?q=an:0393.03047).


*Henson, C. Ward; Moore, L. C. jun.*, [**The nonstandard theory of topological vector spaces**](http://dx.doi.org/10.2307/1996360), Trans. Am. Math. Soc. 172(1972), 405-435 (1973). [ZBL0254.46001](https://zbmath.org/?q=an:0254.46001).


[**Chapter III Nonstandard Theory of Topological Spaces**](https://www.sciencedirect.com/science/article/pii/S0079816908611678)



What's the size of non standard monad for weak topology?
Let's try this, as a negative answer. It has been a long time since I seriously worked on non-standard analysis—so criticism is welcome. I follow mostly the terminology of Robinson's book *Non-Standard Analysis*.


We will assume a [$2^{\aleph\_0}$ saturated model](https://en.wikipedia.org/wiki/Nonstandard_analysis#%CE%BA-saturation).


We consider the Hilbert space $l\_2$. If $a \in l\_2$, write $a(1),a(2),\dots$ for the coordinates of $a$. Let $(e\_n)$ be the unit vectors defined by $e\_n(n) = 1$, $e\_n(k) = 0$ for $k \ne n$.


For $a \in l\_2$, we have $\lim\_{n\to\infty}a(n) = 0$, so
$$
(\forall \epsilon \in \mathbb R, \epsilon > 0)
(\exists n \in \mathbb N)(\forall m \in \mathbb N, m \ge n)\; |a(m)| < \epsilon.
\tag{$1$}$$
By the transfer principle,
$$
(\forall \epsilon \in {}^\*\mathbb R, \epsilon > 0)
(\exists n \in {}^\*\mathbb N)(\forall m \in {}^\*\mathbb N, m \ge n)\; |a(m)| < \epsilon.
\tag{$1'$}$$
Note that we use the same symbol $a$ for the element of ${}^\*l\_2$ corresponding to the original element $a \in l\_2$.  

Now, given $a \in l\_2$ and $\alpha \in {}^\*\mathbb R, \alpha>0, \alpha\simeq 0$,
we have [by $(1')$ with $\epsilon = \alpha^2$] that $E\_a \ne \varnothing$, where
$$
E\_a := \{n \in {}^\*\mathbb N \;:\;(\forall m \in {}^\*\mathbb N, m \ge n)\;|a(m)| < \alpha^2\} .
\tag2$$
Here $\alpha$ is fixed. For each $a \in l\_2$ we have an [internal set](https://en.wikipedia.org/wiki/Nonstandard_analysis#Internal_sets) $E\_a$
because it is defined by the formula $(2)$ in the appropriate language. We claim the family $\{E\_a : a \in l\_2\}$ satisfies the f.i.p. (finite intersection property). Indeed, if $a,b \in l\_2$, then
$c$ defined by $c(n) = |a(n)|+|b(n)|$ for $n \in \mathbb N$ belongs to $l\_2$ and
$$
E\_a \cap E\_b \supseteq E\_c .
$$
The cardinal of $l\_2$ is $2^{\aleph\_0}$. By the $2^{\aleph\_0}$ saturated assumption, we get
$$
E := \bigcap\_{a \in l\_2} E\_a \ne \varnothing.
\tag3$$
Now choose $p \in E$. So $p \in {}^\*\mathbb N$ is an infinite integer.
Let $f \in {}^\*l\_2$ be defined by $f := \alpha^{-1}e\_p$. We claim $f \in \mu(0)$, the weak monad of $0$, but $\|f\|\_2 = \alpha^{-1}$ is infinite.


Proof for $f \in \mu(0)$. Let $a \in l\_2$. Then $p \in E\_a$ and
$$
\big|\langle f,a \rangle\big| = \big|\alpha^{-1}a(p)\big| < \alpha^{-1}\alpha^2 = \alpha \simeq 0 .
$$
This is true for all $a \in l\_2$, so $f \in \mu(0)$.


