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Please help with my understanding of the universal quantifier and its bounded version; You're feedback would be of utmost value.

I'm at discrete math stage, so my context is the day to day math.


Please note that I'm not mentioning any "universal set" here which gives contradictions, but just "a domain of discourse" at hand, like $\mathbb{R}.$


Firstly, Whenever we introduce a variable, it always stands for elements of the domain of discourse in which we are interested, also called as the universe of discourse.


Given a domain of discourse $D$

$\forall x, \phi(x)$ means for every $x$ in our domain of discourse, $\phi(x)$ holds.

Now, from our domain of discourse $D$, we can form subsets. Say, $A \subset U,$ in order to specifically talk about the elements of $A.$

The statement $\forall x \in A, Q(x)$ simply means for all elements $x$ in $A$, $Q(x)$

But because $A \subset D,$ therefore the above statement can always be rephrased equivalently as "for all elements $x$ in our domain of discourse, if $x$ in $A,$ then Q(x)."

Symbolically, we have:

$$\forall x \in A, Q(x) \iff \forall x (x \in A \implies Q(x))$$


Example, Consider Domain of Discourse to be $\mathbb{N}$ and $P$ to be the set of primes. We have $P \subset N.$

As domain is understood, to make a statement about all naturals, we simply write $\forall x, N(x)$

Statement about all primes can be made by restricting ourselves solely to the set of all primes, or by talking of all the naturals which are primes, as both have (convey) the same meaning. Symbolically, $$\forall x \in P, R(x) \iff \forall x (x \in P \implies R(x))$$

"For all primes $x$, $R(x)$" is equivalent to "For all naturals $x$, if $x$ is prime, then $R(x)$."

I would appreciate your time and help.

medium_o
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    Everything you've written is correct, apart from typos. I am not sure what your question is. Is it, "is my understanding correct?"? If so, the answer is "yes". – Mees de Vries Oct 07 '22 at 07:47
  • @MeesdeVries Wow! I'm glad. Thank you very much ! – medium_o Oct 07 '22 at 07:51
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    One thing that often trips up students is the seeming asymmetry between bounded existential and bounded universal quantifiers: we translate $$\forall x\in A, P(x)$$ as $$\forall x(x\in A\color{red}{\rightarrow} P(x))$$ but we translate $$\exists x\in A, P(x)$$ as $$\exists x(x\in A\color{red}{\wedge}P(x)).$$ I've written an explanation of this phenomenon, and why it's not actually an asymmetry, here. – Noah Schweber Oct 07 '22 at 16:40
  • @NoahSchweber Thanks. I must say, in the contexts I come across, your answers have been quite helpful, clear and indeed deepen the understanding. An Esteemed Contributor! – medium_o Oct 07 '22 at 17:08

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