I wanted to apply nonstandard analysis techniques to topology theory, but got trouble finding the equivalent condition to the openness.
It would take too much time to explain all definitions from the begining. Here is a brief definition.
$M$ is a set such that
$\begin{align}&\text{(1)$X\in M\to \mathscr P(X)\in M$ .}\\&\text{(2)$X\in M\to X\subseteq M$}\\&\text{(3)$X,Y\in M\to X\cup Y,X\cap Y\in M$}\end{align}$
$\mathcal L$ is a language whose constant symbols consists of all elements of $M$ , and $\in$ is its only predicate symbol.
The definition of a $\mathcal L$-formula is same as usual, except for that quantifications are required to be of form $(\forall x\in y)$ or $(\exists x\in y)$, and quantifications like $(\forall x)$ or $(\exists x)$ are not allowed here.
$M^*$ is a set that satisfies the following conditions:
$\begin{align}&\text{(1)for every $\mathcal L$-sentence $\varphi$, $M\models\varphi$ iff $M^*\models\varphi$}.\\&\text{(2)for every $X\in M$ and every upward directed partial order $R$ on $X$,}\\&\text{ there is a $y\in X^*$ such that $x^*R^*y$ holds for every $x\in X$.}\\&\text{(3)for every $\mathcal L$-formula $\varphi(x)$ and $X\in M^*$, $\{x\in X:\varphi(x)\}\in M^*$.}\end{align}$
Write $x^*$ for the corresponding element in $M^*$ to $x\in M$.
Now assume $X,Y\in M$ are topological spaces and $\mathscr O_X,\mathscr O_Y\in M$ are their topologies.
For $x\in X$, define $\mu(x)=\bigcap_{x\in O\in \mathscr O_X}O^*$. Pay attention that $\mu(x)\in M^*$ may not hold. Define $\mu(y)$ for $y\in Y$ in the same way.
First, as an example, I will show that $f:X\to Y$ is continuous iff $f^*(\mu(x))\subseteq \mu(f^*(x^*))$ for every $x\in X$.
$\Rightarrow$
Let $f$ be continuous. $f^*(\mu(x))=f^*(\bigcap_{x\in O\in\mathscr O_X}O^*)\subseteq f^*(\bigcap_{f(x)\in O\in \mathscr O_Y}f^{-1^*}(O^*))\subseteq \bigcap_{f(x)\in O\in \mathscr O_Y}f^*(f^{-1^*}(O^*))=\bigcap_{f(x)\in O\in \mathscr O_Y}O^*=\mu(f^*(x^*))$
$\Leftarrow$
Let $O\in\mathscr O_Y$ and $f(x)\in O$. Since all open neighbor hoods of $x$ is upward directed under $\supseteq $, there is some $U\in \mathscr O_X^*$ such that $x^*\in U\subseteq \mu(x)$ holds. Therefore, $M^*\models (\exists U\in \mathscr O_X)(x\in U\wedge f(U)\subseteq O)$ and $M\models (\exists U\in \mathscr O_X)(x\in U\wedge f(U)\subseteq O)$ as well. This shows that $f^{-1}(O)$ is open.
However, when it comes to openess, $\Rightarrow$ can not be shown in the same way, since $\bigcap_{x\in O\in \mathscr O_X} f^*(O^*)\subseteq f^*(\bigcap_{x\in O\in\mathscr O_X}O^*)$ is not necessarily true.
I have neither found a proof for this nor a counter example.