empty set

[lookagain]

Everyone, here is a site that offers three brief paragraph proofs (totalling 13 lines)
on one page for showing that the empty set is a subset of every set. It is from the University of Regina.

The first way looks to be a direct proof. The second way is by intersection of
sets. The third is by contradiction.

1) Type in Mathcentral.uregina.ca

2) Click Math Central [English]

3) Under "Quandaries & queries" type "The empty set is a subset of every set"
in the yellow box. You should leave off the quotation marks.

4) Enter

5) Click on the needed link where you see it lower on that page.

6) The pertinent part begins with the bold "the empty set is a subset of
every set." Please continue reading to the bottom of the page.

----------------------------------------------------------------------------‐-------------


chrislav, by the way, the plural of "monkey" is "monkeys."

 

[chrislav]

I am trying to understand, not argue. I have neither text. Perhaps if you could send me a picture of the Halmos page via PM, I would understand your proof. (It's fair use.)

You say



Now I could restate this in English by choosing 2 as an instance of x.

(1) If 2 is an element of the empty set, then 2 is an element of every set.
(2) Therefore, the empty set is a subset of every set

I agree that statement 1 is true. What I do not see as obvious is how statement 2 follows from statement 1. It is not even the same consequent, and it is certainly not based on modus ponens.

If I put what I interpret that you are saying into modus ponens form,

[math]p \implies q\\ \neg p\\ \therefore q\\ \therefore r.[/math]
That is not a valid argument. Something is missing.

Now after looking at stackexchange, I have come up with this. What I think is missing is an undisclosed generalization step and a specification of how mathematical implication is defined. Let's start with the latter.

[math]\{p \implies q\} \iff \{\neg (p \land \neg q)\} \iff \{(\neg p) \lor q\}.[/math]
That definition of implication justifies the truth of the proposition that a falsity validly implies anything. (It of course does not fit the common meaning of "implication" of English.)

Now in this case we have a definition that depends on implication.

[math]\mathbb A \text { is a subset of } \mathbb B \iff \{x \in \mathbb A \implies x \in \mathbb B\}.[/math]
So we can rewrite the definition as

[math]\mathbb A \text { is a subset of } \mathbb B \iff \{\neg (x \in \mathbb A) \lor (x \in \mathbb B)\}.[/math]
[math]\text {Let } \mathbb B \text { be an ARBITRARY set.}[/math]
[math]\{ \neg (x \in \emptyset) \lor (x \in \mathbb B)\}.[/math]
That statement is necessarily true whether or not x is an element of set B. The term x "cancels out" as it were. The x is definitely not in the empty set.

[math]\therefore \text {BY DEFINITION, } \emptyset \subset \mathbb B.[/math]
But B was an arbitrary set. Therefore the empty set is a subset of any set.

I think that proof is valid. And I think that is what pka's proof meant. But whether that proof is simple in terms of being obvious is a judgment about the audience.

[math]\{ \neg (x \in \emptyset) \lor (x \in \mathbb B)\}.[/math]You are just a step away by giving a direct proof
use that law :[math]p\vee q\Leftrightarrow\neg p\implies q[/math]And then double negation rule
And then of course Universal Introduction
Where p is:[math]x\in\emptyset[/math]and q is :[math]x\in B[/math]

 

[chrislav]

Everyone, here is a site that offers three brief paragraph proofs (totalling 13 lines)
on one page for showing that the empty set is a subset of every set. It is from the University of Regina.

The first way looks to be a direct proof. The second way is by intersection of
sets. The third is by contradiction.

1) Type in Mathcentral.uregina.ca

2) Click Math Central [English]

3) Under "Quandaries & queries" type "The empty set is a subset of every set"
in the yellow box. You should leave off the quotation marks.

4) Enter

5) Click on the needed link where you see it lower on that page.

6) The pertinent part begins with the bold "the empty set is a subset of
every set." Please continue reading to the bottom of the page.

----------------------------------------------------------------------------‐-------------


chrislav, by the way, the plural of "monkey" is "monkeys."

Those are monkies from the moon

 

[JeffM]

[math]\{ \neg (x \in \emptyset) \lor (x \in \mathbb B)\}.[/math]You are just a step away by giving a direct proof
use that law :[math]p\vee q\Leftrightarrow\neg p\implies q[/math]And then double negation rule
And then of course Universal Introduction
Where p is:[math]x\in\emptyset[/math]and q is :[math]x\in B[/math]

Dude

I am not trying to convince you. I am trying to understand an argument. What is “Universal Introduction”?

 

[chrislav]

@Jeff, all I can say is here are two refferences.
NAIVE SET THEORY by Paul Halmos page 6.
ELEMENTS OF SET THEOR by Enderton pp 2-6

Paul Halmos mentions the vacuously truth "proof" and then gives the usual proof by contradiction
But we will discuss in another thread the value of that book

 

[JeffM]

As I conceded before, I own neither text cited. Given that I am now in quarantine, it is difficult to consult either. Thus, I find it difficult to grasp what you intend to convey by repeating a post that I already explained is totally uninformative as far as I am concerned.

Nevertheless, please do try again to educate me.

 

[chrislav]

Dude

I am not trying to convince you. I am trying to understand an argument. What is “Universal Introduction”?

Definitionuniversal introduction)
From a wwf [math]\phi[/math]containing a name letter 'a' not occuring in any assumption or in any hypothesis in effect at the line on which [math]\phi[/math] occurs,we may infer any wwf of the form[math]\forall\beta\phi_ \beta/a[/math] where [math]\phi_\beta/a[/math] is the result of replacing all occurancesof a in [math]\phi[/math] by some variable [math]\beta[/math] not alleady in [math]\phi[/math]I will explain in more details tomorrow and how this law can be aplied in our case

 

[JeffM]

wwf = world wildlife fund?

I am definitely not getting this proof. I thought I had groked it, but clearly I was wrong. Think I shall take a break from this site for a while.

 

[lookagain]

give a direct proof that the empty set is a subset of every set

chrislav, I was caught off guard because you did not bother to write in a sentence.

If this was intended to be a "challenge problem," where you supposedly know the
solution, but you want the other members to post to see how their alleged solutions
look, then it is a must that you indicate that with your thread post so they know
from where you are coming.
Else, as the comment from post # 3 indicated, you would show some work (and/or
pertinent questions) along with the question in your first post.

 

[lookagain]

wwf = world wildlife fund?

I am definitely not getting this proof. I thought I had groked it, but clearly I was wrong. Think I shall take a break from this site for a while.

JeffM, did you check out the proofs at the site in my post # 21 yet? The OP's username is "chrislav," not "Dude."

 

[JeffM]

Did you check out the proofs at the site in my post # 21 yet?

@lookagain

I was not doubting the result. Nor was I demanding an answer that might be intuitive to me although I would have been appreciative of one. And I appreciate that your providing alternative proofs may very well be what will help the OP. My post had a different thrust.

I was trying to understand the proof asserted to be ”quite easy.”. I did not find it so. I suggested that something was missing. Rather than given any attempt to clarify the argument, I was then dismissed with an argument by authority (although I concede that it is my own problem that the authorities referred to are not in a library currently available to me). Instead, I spent quite some time at math stack exchange looking for similar arguments and trying to understand them.

Thus, I apologize for not looking at your alternative proofs. I am relatively confident that I now understand the logic of the proof I asked about. And obviously alternative proofs cannot help me grasp pka’s two line proof.

In another recent thread, I said that what was sufficient as a proof is subjective. My only problem in this thread, and my only reason for interjecting myself into it, is that when the OP did not understand the proof given, he was told that “you simply do not know enough about this subject.” I had hoped, incorrectly it seems, that posing the question in a different way might lead to a fuller and therefore more helpful explanation. Instead, we have been told that asking questions demonstrates that we do not know enough about the subject to be considered worthy of a meaninful answer and should read some books that are assumed to be available to us.

You and I have argued in the past. I do not like to argue. I appreciate that you have tried to address the substantive issue raised by the OP. I do not want to argue with you about the validity of alternative proofs. My sole concern here has been how to articulate more comprehensibly an argument that I assumed was valid but excessively truncated. I am not sure I have done so. Any critique of my version of what I think pka meant will be welcomed.