I am trying to express the definition of a "stable" point using quantifiers. This is a homework problem from a course I am following, but not taking. (Edited very lightly, mainly to simplify notation.)
We say that an equilibrium $x$ is stable if for each open set $\mathcal{U}$ containing $x$ there exists a smaller open set $\mathcal{V} \subset \mathcal{U}$ containing $
x$ such that for all $b \in \mathcal{V}$, we have $\varphi(t,b \in \mathcal{U}$ for all $t \geq
0$. Give a careful definition of what it means for an equilibrium to be unstable. Don't just say that
"unstable means not stable"! Instead, give an unabridged definition that uses the quantifiers "there
exists" and "for all."
I tried writing down the definition of stable and negating it, but I produced something rather messy that was either incorrect or not in fact the negation of the original statement. So my attempt to negate the statement is:
\begin{align*}
\exists \mathcal{U}, \; x \in \mathcal{U} \wedge \left(\mathcal{V} \subset \mathcal{U}, x \in \mathcal{V} \implies \; \exists b \in \mathcal{V}, \; \exists t \geq 0, \; \varphi(t, b) \not \in \mathcal{U} \right).
\end{align*}
So, in effect, I'm saying I can find an open set $U$ which contains $x$ and, no matter which subset $V \subset U$ I pick which contains $U$, I can find a $b \in V$ and a $t \geq 0$ so that $\varphi(t,b) \not \in \mathcal{U}$.
I don't have a predicate for expressing that a set is open, so I believe I am taking as a convention that the "mathcal" environment denotes an open set.
My attempt at directly expressing this statement is:
\begin{align*}
\forall \mathcal{U}, \; x \in \mathcal{U} \implies \left(\exists \mathcal{V}, \; \mathcal{V} \subset U \wedge x \in \mathcal{V} \wedge \left(\forall b \in \mathcal{V}, \forall t \geq 0, \varphi(t,b) \in \mathcal{U} \right) \right).
\end{align*}
I don't see a way to negate this to the above definition I wrote down directly, so I must have made a mistake on at least one of them.
How does this look? I'm hoping to understand how to quantify both the definition and its negation without defining additional predicates.
Best Answer
About quantifier scope:
With reference to your translation of the definition: this formula (what you write) $$(\forall y, P\land Q)$$ may actually mean $$\forall y P\land Q\quad\text{i.e.,}\quad(\forall y\, P)\land Q,$$ instead of your intended $$\forall y\,(P\land Q).$$
The original statement (where calligraphic font denotes open sets): $$\forall \mathcal{U} \bigg( \; x \in \mathcal{U} \to\exists \mathcal{V}\bigg( \; \mathcal{V} \subset \mathcal U \wedge x \in \mathcal{V} \wedge \forall b \in \mathcal{V} \,\forall t \geq 0 \,\varphi(t,b) \in \mathcal{U} \bigg)\bigg).$$
Negating it into Prenex normal form: $$\exists\mathcal{U}\bigg( \; x \in \mathcal{U} \land \lnot\exists \mathcal{V}\bigg( \; \mathcal{V} \subset \mathcal U \wedge x \in \mathcal{V} \wedge \forall b \in \mathcal{V} \,\forall t \geq 0 \,\varphi(t,b) \in \mathcal{U} \bigg)\bigg) \\\equiv \exists\mathcal{U}\bigg( \; x \in \mathcal{U} \land \forall\mathcal{V}\bigg( \; \mathcal{V} \not\subset \mathcal U \lor x \not\in \mathcal{V} \lor \exists b \in \mathcal{V} \,\exists t \geq 0 \,\varphi(t,b) \not\in \mathcal{U} \bigg)\bigg) \\\equiv \exists\mathcal{U}\bigg( \; x \in \mathcal{U} \land \forall\mathcal{V}\,\exists b \in \mathcal{V} \,\exists t \geq 0 \bigg( \; \mathcal{V} \not\subset \mathcal U \lor x \not\in \mathcal{V} \lor \varphi(t,b) \not\in \mathcal{U} \bigg)\bigg) \\\equiv \exists\mathcal{U}\,\forall\mathcal{V}\,\exists b \in \mathcal{V} \,\exists t \geq 0 \bigg( \; x \in \mathcal{U} \land \bigg( \; \mathcal{V} \not\subset \mathcal U \lor x \not\in \mathcal{V} \lor \varphi(t,b) \not\in \mathcal{U} \bigg)\bigg).$$
(Alternatively—and slightly more efficiently—convert to Prenex form before negating.)
Your attempt is almost equivalent to the second line of my negation, only missing the $\forall\mathcal{V}.$