Hello everyone. In this video, we're going to talk about the following question. This is going to be an example of nested quantifiers and negating them. I haven't done too many examples of applications of what we've learned, so I'm going to do an application in calculus today, and it's going to be with limits.
Express $\displaystyle\lim_{x\to a} f(x) \neq L$ in terms of predicate logic.
So first thing, what is predicate logic? Predicate logic, it means all those symbols that we've been using in the course, so $\forall$, $\exists$, $\Rightarrow$, $\land$, $\lor$, $\neg$ etc.
To get this started, so this is a "$\neq$"" question, but in this case it's far easier to write down the equality and then go from there.
Now let me mention something about this problem before I go too far; if you don't know the calculus behind this, that's okay. There's still something to learn here, but if you'd like you can zoom ahead a little bit and then just look at the negating of the nested quantifiers and the actual writing out of the English sentence; those are the two most important things.
Okay so what is this in English?
For every $\epsilon >0$, there exists a $\delta>0$ such that for each real number $x$ where $|x-a|<\delta$ holds, then $|f(x)-L|<\epsilon$.
So here's the sentence, so this is what the equality limit means in English, there's an "if", so you should be thinking implication.
Okay so let's do this now, so let's turn this into a predicate logic statement.
$$\forall\epsilon >0\; \exists\delta >0\; \forall x \in \mathbb{R}(|x-a|<\delta \Rightarrow|f(x)-L|<\epsilon).$$So I just want to mention a couple of notations. Instead of $\forall\epsilon >0$, you might see $\forall\epsilon\in\mathbb{R}^+$, which means for all $\epsilon$ in the positive reals. You might see this notation, $\forall\epsilon\in\mathbb{R}_{>0}$, this also means for all $\epsilon$ in the positive reals.
I'm not going to specify the domain here just because I haven't done an example of this in class and I want to do one. But you can write this in many different ways, so I'm going to write it as $\forall\epsilon >0$
So "there exists a $\delta>0$ such that for each real number $x$ where $|x-a|<\delta$ holds, then $|f(x)-L|<\epsilon$" is a hidden "if statement" so it's good that I did this just so that you get another example.
So since I'm saying for each real number $x$ that's the same as "for all".
So remember intuitively what this says, right, it says that $f(x)$ gets infinitely closer to $L$, as $x$ approaches $a$, as long as I can pick a neighbourhood near $a$, where as long as I'm close enough to $a$, then the function is close enough to $L$. That's really the intuitive definition here, but that's not relevant for algebra.
So here's a predicate logic statement for the equality. Now we have to negate this. Again the way to think about it is that, if I'm negating a 'for all' statement then it should be a 'there exists' statement, and if I'm negating a 'there exists' statement it should be a 'for all' statement.
So let's pound through this, and then we get to this implication part at the end, so let's look at that in a second.
$$\begin{align*} \neg&(\forall\epsilon >0\; \exists\delta >0\; \forall x \in \mathbb{R}(|x-a|<\delta \Rightarrow|f(x)-L|<\epsilon))\\\\ &\equiv \exists\epsilon >0\;\forall\delta >0\;\exists x \in \mathbb{R}\neg(|x-a|<\delta \Rightarrow|f(x)-L|<\epsilon) \end{align*}$$Notice that the domains don't change when we negate. Now we're going to get the negation of the implication. What's the negation of an implication? Well we can do a little aside to discover:
$$\begin{align*} \neg(A\Rightarrow B)&\equiv \neg(\neg A \lor B)\\ &\equiv A \land \neg B \end{align*}$$So $\neg(A\Rightarrow B)\equiv A \land \neg B$, so let's just take $A$ and $\neg B$ here, where $A=|x-a|<\delta$ and $\neg B=\neg(|f(x)-L|<\epsilon)\equiv |f(x)-L|\geq\epsilon$
$$\begin{align*} \neg&(\forall\epsilon >0\; \exists\delta >0\; \forall x \in \mathbb{R}(|x-a|<\delta \Rightarrow|f(x)-L|<\epsilon))\\ &\equiv \exists\epsilon >0\;\forall\delta >0\;\exists x \in \mathbb{R}\neg(|x-a|<\delta \Rightarrow|f(x)-L|<\epsilon)\\ &\equiv \exists\epsilon >0\;\forall\delta >0\;\exists x \in \mathbb{R}(|x-a|<\delta \land|f(x)-L|\geq\epsilon) \end{align*}$$And here we have it. So here's an example of the following question: express $\displaystyle\lim_{x\to a} f(x) \neq L$ in terms of predicate logic.
So how do we do this? Well first we express the equality in terms of English, then in terms of predicate logic, and then we negated it. Sometimes it's hard to just go straight from question to answer. Sometimes there's a bunch of intermediate steps that will help you get there.
I hope this helps with understanding nested quantifiers and with negating them. Again, maybe I'll make one more mention now since I'm here, so again, $\forall\epsilon >0$, so think about it like an adversary game, right?
So your adversary picks an $\epsilon >0$, you get to pick a $\delta$ that may or may not depend on $\epsilon$, then your adversary says, "Okay well for every $x$ such that you're inside this nice little area here, so you're inside this nice neighbourhood around $a$, then you must have your function sufficiently close to your limit. If this is to be equal."
You could think about this game all the time, right, so you can think of the $\forall$ is sort of like the enemy picks, and an $\epsilon$ that you don't know, you can think of your $\exists$ as you get a little bit of control as to thinking about which $\delta$ you want. Alright so hopefully this video helps. Thank you very much for listening.