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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.

ryang
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Mathematical Rookie
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  • It would be helpful if you could provide a bit more context: specifically: why don't you have a predicate for expressing that a set is open?; is it an exercise from a book or a course or something you have come up with? If it is intended to be implicit that sets written in calligraphic fonts are open sets, your second attempt looks pretty good to me (apart from a $U$ that should be $\cal U$). Why do you feel the need to negate anything? – Rob Arthan Sep 06 '21 at 20:30
  • It's an exercise from a course I'm following but not taking, yes. I don't believe it's standard convention in any book I've seen that calligraphic letters are open sets, but I believe that was the intention of the author of the problem. And, yes, the problem asks for a negation. Thanks for the feedback on the second quantification. Does the first look ok as well? – Mathematical Rookie Sep 06 '21 at 20:35
  • It would be a good idea to quote the problem in more detail, including the requirement to come up with a negation. $\forall x \ni {\cal V} \subset {\cal U}$ in your attempt at the negation doesn't look right: I'd have another think about that: try to more systematic about carrying out the negation. And, if you want to give information in response to a comment, then please edit your question to include that information. – Rob Arthan Sep 06 '21 at 20:41
  • @Rob Arthan: Ok, let me quote from the problem directly in that case. I'll make that revision now. – Mathematical Rookie Sep 06 '21 at 20:48
  • @RobArthan Ok, I edited the question and will take another look at my attempted negation. – Mathematical Rookie Sep 06 '21 at 20:51
  • I think the notation $\forall x\ni\mathcal V\subset\mathcal U$ is confusing because it looks as if you're quantifying $x$ ("for all $x\ \dots$"), when your intention is to quantify $\mathcal V$. It would be better to write explicitly $\forall\mathcal V,(x\in\mathcal V\subseteq U \implies \dots)$. – Andreas Blass Sep 06 '21 at 20:57
  • @AndreasBlass I edited my attempt. Does this look better? I agree that I made an error in appearing to quantify over $x$. – Mathematical Rookie Sep 06 '21 at 21:04

1 Answers1

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  1. About quantifier scope:

    • a comma may not clearly indicate the scope of quantification;
    • adding parentheses around a quantifier may add no information.

    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 $$\forall y\,(P\land Q).$$

  2. The original statement (where calligraphic font denotes open sets): $$\forall \mathcal{U} \bigg( \; x \in \mathcal{U} \to\exists \mathcal{V}\bigg( \; \mathcal{V} \subset \mathcal U \wedge x \in \mathcal{V} \wedge \forall b \in \mathcal{V} \,\forall t \geq 0 \,\varphi(t,b) \in \mathcal{U} \bigg)\bigg).$$

  3. 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 $\forall\mathcal{V}.$

ryang
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  • Everything up to the second line of negation makes total sense to me, but I can't fiugure out out how you converted it to the third line and moved the "$\exists b$" forward in the quantification. Could you say a bit more about where that came from? – Mathematical Rookie Sep 07 '21 at 03:16
  • @MathematicalRookie I was combining two ideas, which are made clearer in this and this previous answers that I wrote. Hopefully I haven't made any mistake! I've also just added a Point $0$ to the answer. – ryang Sep 07 '21 at 06:14
  • @MathematicalRookie The first link in the previous comment no longer works as its parent Question has been deleted; refer to the bottom of this Answer instead and the embedded link within. – ryang Sep 12 '21 at 07:34