MHB What does "v belongs to v" mean?

[evinda]

Hello! (Wave)

In order to show that not every collection of objects is a set, we do the following:

We will show that if we suppose that "everything is a set", we are lead to contradictions.

If $V:=\{ x: x \notin x \}= \text{ the collection of all objects, that don't belong to themselves }$ , then $V$ is a set.

Therefore, we conclude that:

$$v \in V \Leftrightarrow v \notin v, \text{ contradiction.}$$

But.. what does $v \in v$ mean?

 

[I like Serena]

Hello! (Wave)

In order to show that not every collection of objects is a set, we do the following:

We will show that if we suppose that "everything is a set", we are lead to contradictions.

If $V:=\{ x: x \notin x \}= \text{ the collection of all objects, that don't belong to themselves }$ , then $V$ is a set.

Therefore, we conclude that:

$$v \in V \Leftrightarrow v \notin v, \text{ contradiction.}$$

But.. what does $v \in v$ mean?

Hi! (Smile)

Let's see...

If we say $a \in b$, then that means that $b$ is of the form $b=\{ a, ... \}$.
So if $a \in a$, then that means that $a = \{ a, ... \}$.
In other words $a$ is both a set and an element.

As an example, consider the set:
$$P = \{ \varnothing, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \}, ...\}$$
See how for instance $\{ \varnothing \}$ is an element of $P$?

Hmm... would $P \in P$? (Wondering)

 

[evinda]

Let's see...

If we say $a \in b$, then that means that $b$ is of the form $b=\{ a, ... \}$.
So if $a \in a$, then that means that $a = \{ a, ... \}$.
In other words $a$ is both a set and an element.

A ok.. (Nod)

As an example, consider the set:
$$P = \{ \varnothing, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \}, ...\}$$
See how for instance $\{ \varnothing \}$ is an element of $P$?

Hmm... would $P \in P$? (Wondering)

Yes, I see.. So that $P \in P$ is satisfied, shouldn't it be like that:

$$P = \{P, \varnothing, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \}, ...\}$$

Or am I wrong? (Sweating)

If $V:=\{ x: x \notin x \}= \text{ the collection of all objects, that don't belong to themselves }$ , then $V$ is a set.

Therefore, we conclude that:

$$v \in V \Leftrightarrow v \notin v, \text{ contradiction.}$$

Why is this then a contradiction?

 

[Evgeny.Makarov]

If $V:=\{ x: x \notin x \}= \text{ the collection of all objects, that don't belong to themselves }$ , then $V$ is a set.

Therefore, we conclude that:

$$v \in V \Leftrightarrow v \notin v, \text{ contradiction.}$$

In the last line, all $V$'s should be capital:
\[
V\in V\iff V\notin V.\qquad(*)
\]
This is indeed a contradiction because $V\notin V$ means $\neg(V\in V)$. The statement (*) is a special case of the fact that if $X=\{x: P(x)\}$, then $y\in X\iff P(y)$. This is the definition of the set-builder notation.

Edit: The equivalence
\[
v \in V \Leftrightarrow v \notin v
\]
is also true, but it is not a contradiction yet. It becomes one when $v$ is instantiated with $V$.

 

[Deveno]

I think the contradiction should be:

$V \in V \iff V \not\in V$.

The reason I say this, is if:

$V = \emptyset = \{\ \}$

then $v \in V$ never happens, and there is no contradiction.

The way I like to phrase this is: there IS a Russell set, but it's empty.

********************

The origins of this line of thought are very old. If one supposes that statements must either be true, or false (the law of the excluded middle), then a statement such as:

"This statement is false"

presents us with a quandry. This state of affairs is known as "The Liar's Paradox". Assuming the statement is true, leads us to assert it is false, and vice versa.

Here is a slightly more sophisticated version:

A word is called "autologic" if it refers to itself. Examples are: "English", and "seventeenlettered". A word is called "heterologic" if it does NOT refer to itself: examples are "French" and "threelettered".

One then can ask: is "heterologic" heterologic, or autologic?

If, for a non-empty set, we could have $a \in a$, it becomes problemmatic to describe the elements of $a$.

To see this, suppose $x \in a$.

Then a partial listing of $a$ is as follows:

$a = \{x,\{x,\{x,\{x\dots$

Clearly, we're never going to be able to finish "writing down $a$". This is kind of a problem.

However, we want to be able to do "something like this", for example, many functions (and indeed the natural numbers themselves) are defined by RECURSION, in which self-reference is unavoidable.

One way to avoid this kind of thing is to insist sets are "well-founded" (this is the approach taken by Zermelo-Fraenkel) and "well-defined". In other words, not just ANYTHING can be a set, only CERTAIN things can be sets, but once we HAVE a bona-fide set, any way of specifying PART of it (restricted comprehension) is OK.

Stating this in a technical and precise way, is part of what makes set theory "so abstract", because we have to do so in way that is (we hope) logically CONSISTENT. This is harder than it looks-many of our "intuitive" ways of reasoning about things can get us into trouble applied "willy-nilly". This is particularly true when reasoning about INFINITE things, which unfortunately, is MOST of mathematics.

A "naive" form (somewhat weaker) of Russell's paradox (the set of all sets $R = \{x: x\not\in x\}$) is called "who shaves the barber":

In a certain town, it is decreed that all men must shave, and furthermore, that all men (and only those men) who do not shave themselves, MUST be shaved by the barber. So-who shaves the barber?

Some mathematicians think the law of the excluded middle is too strong, that insisting either something is, or is not the case, is too sweeping. These people have three categories:

Proven to be true.
Proven not to be true.
None of the above.

Moreover, the methods of proof they will accept, are likewise limited:

Able to be proven (constructible)
Able to be disproven (counter-example constructible)
Proof or disproof unknown

In particular, proofs by contradiction are generally not allowed. It should be noted that a good deal of "classical mathematics" can be verified by these means, leading one to suspect that the mathematical universe in this view isn't all "that much smaller".

 

[evinda]

In the last line, all $V$'s should be capital:
\[
V\in V\iff V\notin V.\qquad(*)
\]
This is indeed a contradiction because $V\notin V$ means $\neg(V\in V)$. The statement (*) is a special case of the fact that if $X=\{x: P(x)\}$, then $y\in X\iff P(y)$. This is the definition of the set-builder notation.

Edit: The equivalence
\[
v \in V \Leftrightarrow v \notin v
\]
is also true, but it is not a contradiction yet. It becomes one when $v$ is instantiated with $V$.

So, we have $V \in V$, since $V$ is a set and each set belongs to itself?
Also, do we get that $V \notin V$, since each element $x \in V$, satisfies the property $x \notin V$ ? (Thinking)

 

[Evgeny.Makarov]

So, we have $V \in V$, since $V$ is a set and each set belongs to itself?
Also, do we get that $V \notin V$, since each element $x \in V$, satisfies the property $x \notin V$ ?

We have neither $V \in V$ nor $V\notin V$, but we do have $V\in V\iff V\notin V$. The reasons you gave for $V \in V$ and $V\notin V$ don't make sense to me, either. For example, usually the phrase "$x$ belongs to $A$" means $x\in A$ (and not, say, $x\subseteq A$). In this sense, no set belongs to itself.

The reason $V\in V\iff V\notin V$ holds, as I said in post #4, is the definition of the set-builder notation and the definition of $V$.

 

[evinda]

We have neither $V \in V$ nor $V\notin V$, but we do have $V\in V\iff V\notin V$. The reasons you gave for $V \in V$ and $V\notin V$ don't make sense to me, either. For example, usually the phrase "$x$ belongs to $A$" means $x\in A$ (and not, say, $x\subseteq A$). In this sense, no set belongs to itself.

The reason $V\in V\iff V\notin V$ holds, as I said in post #4, is the definition of the set-builder notation and the definition of $V$.

In our case, the set $X=\{ x:P(x) \}$ is this one: $V=\{ x: x \notin x\}$.

And we have that $y \in V \Leftrightarrow y \notin x$.

How do we conclude from this, that $V \notin V$? (Worried)

I am confused now... (Sweating)

 

[Evgeny.Makarov]

In our case, the set $X=\{ x:P(x) \}$ is this one: $V=\{ x: x \notin x\}$.

And we have that $y \in V \Leftrightarrow y \notin x$.

No, that is not so. I don't know whether what you wrote is a typo or a major misunderstanding, so I think it makes sense to go over the background. Please forgive me if this seems trivial to you.

Two major operations in mathematics are matching two expressions and instantiation of a variable with an expression. Theorems often have the form "For all $x$, $P(x)$ implies $Q(x)$". To apply this theorem to a concrete situation where $P(E)$ holds, we have first to compare, or match, the premise $P(x)$ of the theorem with our concrete statement $P(E)$. The comparison should yield the value of $x$ in our concrete situation. Suppose that what the theorem calls $x$ is, in our situation, some concrete expression $E$. The next step is instantiating $x$ with $E$ in the conclusion $Q(x)$ to get $Q(E)$. This operation is also called replacing $x$ with $E$ and substituting $E$ for $x$. (Note that order in which $x$ and $E$ are used.) Instantiation means going over $Q(x)$ and every time one sees $x$, replacing it with $E$. This may sound trivial (and it is), but I've had students who had trouble doing it. If you are not sure you fully understand instantiation, you may want to try the following.

• Instantiate $A$ in $\forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow A = B)$ with $\varnothing$.
• Instantiate $x$ in $\forall A \, \exists B \, ( x \in B \Leftrightarrow [ x \in A \land \varphi(x, w , A) ] )$ with $w$.
• Instantiate $B$ in $[B \in P \iff \forall C \, (C \in B \Rightarrow C \in A)]$ with $\{A\}\cup A$.

One complication arises when the expression $E$ you have to substitute has a variable that is quantified or otherwise bound in the expression $P(x)$. For example, $P(x)$ can be $\exists y\;x\le y$ and $E$ can be $y+1$. Here $E$ has $y$, which is quantified in $P(x)$. Then instantiating $x$ with $E$, i.e., replacing $x$ by $E$, produces $\exists y\;y+1\le y$. The original statement $\exists y\;x\le y$ was true for all $x$ (take $y=x$), but the resulting statement $\exists y\;y+1\le y$ is always false, so we did something wrong. To do the substitution correctly, one has first to rename the variable $y$ bound by the quantifier since $\exists y\;x\le y$ means the same as $\exists z\;x\le z$. Then replacing $x$ with $y+1$ gives $\exists z\;y+1\le z$.

Sometimes you have to instantiate variables with functions, properties or relations. For example, in your case the property $P(x)$ has to be instantiated with $x\notin x$. If before instantiation you have, for example, $P(\{y\})$, then replacing $P(x)$ with $x\notin x$ gives $\{y\}\notin\{y\}$. The precise rules become technical and probably won't help an intuitive understanding, but you should be able to do this in concrete cases.

Back to set-builder notation. I said in post #4 that
\[
X=\{x: P(x)\}\implies \forall y\;y\in X\iff P(y).\qquad(*)
\]
This is the definition of the set-builder notation. In your case, $V=\{x:x\notin x\}$. To apply (*) to this set, we compare the premise of (*), i.e., $X=\{x: P(x)\}$, with $V=\{x:x\notin x\}$. Comparing symbol by symbol, we see that $X$ is $V$ and $P(x)$ is $x\notin x$. Now let's put our instantiating skills to use and instantiate $X$ with $V$ and $P(x)$ with $x\notin x$ in the conclusion of (*), i.e., in $\forall y\;y\in X\iff P(y)$. The result is $\forall y\;y\in V\iff y\notin y$. Since this is a true statement starting with $\forall y$, we can instantiate $y$ with anything, for example, with $V$. This gives $V\in V\iff V\notin V$.

How do we conclude from this, that $V \notin V$?

(For the background of this meme, see the second dialog here.)

I say it again, nobody claims that $V\notin V$. What the argument proves is that $V\in V\iff V\notin V$ holds for $V$ with that definition. You know that there are atomic (indivisible) propositions such as $x\in A$ and $(x+y)^2=x^2+2xy+y^2$. There are also compound propositions that are formed from atomic ones using "and", "or", "not", "implies" and other connectives. For example, $x\in A\land (x+y)^2=x^2+2xy+y^2$ is a compound proposition. In this case, the conclusion of the argument is a compound proposition $V\in V\iff V\notin V$ that consists of two atomic propositions $V\in V$ and $V\notin V$ joined by the connective $\iff$, i.e., equivalence.

 

[evinda]

No, that is not so. I don't know whether what you wrote is a typo or a major misunderstanding, so I think it makes sense to go over the background. Please forgive me if this seems trivial to you.

Two major operations in mathematics are matching two expressions and instantiation of a variable with an expression. Theorems often have the form "For all $x$, $P(x)$ implies $Q(x)$". To apply this theorem to a concrete situation where $P(E)$ holds, we have first to compare, or match, the premise $P(x)$ of the theorem with our concrete statement $P(E)$. The comparison should yield the value of $x$ in our concrete situation. Suppose that what the theorem calls $x$ is, in our situation, some concrete expression $E$. The next step is instantiating $x$ with $E$ in the conclusion $Q(x)$ to get $Q(E)$. This operation is also called replacing $x$ with $E$ and substituting $E$ for $x$. (Note that order in which $x$ and $E$ are used.) Instantiation means going over $Q(x)$ and every time one sees $x$, replacing it with $E$. This may sound trivial (and it is), but I've had students who had trouble doing it. If you are not sure you fully understand instantiation, you may want to try the following.

• Instantiate $A$ in $\forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow A = B)$ with $\varnothing$.
• Instantiate $x$ in $\forall A \, \exists B \, ( x \in B \Leftrightarrow [ x \in A \land \varphi(x, w , A) ] )$ with $w$.
• Instantiate $B$ in $[B \in P \iff \forall C \, (C \in B \Rightarrow C \in A)]$ with $\{A\}\cup A$.

So, is it like that?

• $\forall B(\forall x(x \in \varnothing \Leftrightarrow x \in B) \Rightarrow \varnothing=B)$
  $$$$
• $\forall A \exists B(w \in B \Leftrightarrow [w \in A \wedge \phi(w,w',A)])$
  $$$$
• $[\{A\} \cup A \in P \Leftrightarrow \forall C(C \in \{A\} \cup A \Rightarrow C \in D)]$

$$$$

One complication arises when the expression $E$ you have to substitute has a variable that is quantified or otherwise bound in the expression $P(x)$. For example, $P(x)$ can be $\exists y\;x\le y$ and $E$ can be $y+1$. Here $E$ has $y$, which is quantified in $P(x)$. Then instantiating $x$ with $E$, i.e., replacing $x$ by $E$, produces $\exists y\;y+1\le y$. The original statement $\exists y\;x\le y$ was true for all $x$ (take $y=x$), but the resulting statement $\exists y\;y+1\le y$ is always false, so we did something wrong. To do the substitution correctly, one has first to rename the variable $y$ bound by the quantifier since $\exists y\;x\le y$ means the same as $\exists z\;x\le z$. Then replacing $x$ with $y+1$ gives $\exists z\;y+1\le z$.

Sometimes you have to instantiate variables with functions, properties or relations. For example, in your case the property $P(x)$ has to be instantiated with $x\notin x$. If before instantiation you have, for example, $P(\{y\})$, then replacing $P(x)$ with $x\notin x$ gives $\{y\}\notin\{y\}$. The precise rules become technical and probably won't help an intuitive understanding, but you should be able to do this in concrete cases.

Back to set-builder notation. I said in post #4 that
\[
X=\{x: P(x)\}\implies \forall y\;y\in X\iff P(y).\qquad(*)
\]
This is the definition of the set-builder notation. In your case, $V=\{x:x\notin x\}$. To apply (*) to this set, we compare the premise of (*), i.e., $X=\{x: P(x)\}$, with $V=\{x:x\notin x\}$. Comparing symbol by symbol, we see that $X$ is $V$ and $P(x)$ is $x\notin x$. Now let's put our instantiating skills to use and instantiate $X$ with $V$ and $P(x)$ with $x\notin x$ in the conclusion of (*), i.e., in $\forall y\;y\in X\iff P(y)$. The result is $\forall y\;y\in V\iff y\notin y$. Since this is a true statement starting with $\forall y$, we can instantiate $y$ with anything, for example, with $V$. This gives $V\in V\iff V\notin V$.(For the background of this meme, see the second dialog here.)

I say it again, nobody claims that $V\notin V$. What the argument proves is that $V\in V\iff V\notin V$ holds for $V$ with that definition. You know that there are atomic (indivisible) propositions such as $x\in A$ and $(x+y)^2=x^2+2xy+y^2$. There are also compound propositions that are formed from atomic ones using "and", "or", "not", "implies" and other connectives. For example, $x\in A\land (x+y)^2=x^2+2xy+y^2$ is a compound proposition. In this case, the conclusion of the argument is a compound proposition $V\in V\iff V\notin V$ that consists of two atomic propositions $V\in V$ and $V\notin V$ joined by the connective $\iff$, i.e., equivalence.

I understand... (Nod) But, how can we explain that $V \in V \Leftrightarrow V \notin V$ is a contradiction? (Thinking)

 

[Evgeny.Makarov]

Instantiate $A$ in $\forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow A = B)$ with $\varnothing$.

$\forall B(\forall x(x \in \varnothing \Leftrightarrow x \in B) \Rightarrow \varnothing=B)$

Correct.

Instantiate $x$ in $\forall A \, \exists B \, ( x \in B \Leftrightarrow [ x \in A \land \varphi(x, w , A) ] )$ with $w$.

$\forall A \exists B(w \in B \Leftrightarrow [w \in A \wedge \phi(w,w',A)])$

No, the last part should be $\varphi(w,w,A)$ because $w$ is not bound (quantified) in the formula. It's like when you have a formula that has 0 and $x$ and you instantiate $x$ to 0: you don't rename the existing 0.

Instantiate $B$ in $[B \in P \iff \forall C \, (C \in B \Rightarrow C \in A)]$ with $\{A\}\cup A$.

$[\{A\} \cup A \in P \Leftrightarrow \forall C(C \in \{A\} \cup A \Rightarrow C \in D)]$

Again, the last $A$ is not replaced by some unknown $D$. $A$ is a parameter in the original formula, so it should be treated like a constant. The end of the resulting formula should be $C\in A\cup\{A\}\Rightarrow C\in A$.

But, how can we explain that $V \in V \Leftrightarrow V \notin V$ is a contradiction?

Any proposition of the form $P\leftrightarrow\neg P$ is a contradiction. Whether $P$ is true or false, the whole proposition is false.

There is one potential confusion. In logic there are two implications, often denoted by $\implies$ (or its shorter variant $\Rightarrow$) and $\to$, as well as two equivalences: $\iff$ (or $\Leftrightarrow$) and $\leftrightarrow$. The bold arrows are not part of the formal language, i.e., not part of formulas. They are a part of our natural language when we speak about formulas. Thus, $P\implies Q$ means that formula $P$ implies formula $Q$. Similarly, $P\iff Q$ means hat $P$ is equivalent to $Q$. In contrast, $\to$ is a part of the formal language and may occur in formulas. We have, for example, $P\to Q\iff \neg P\lor Q$. This means that the formulas $P\to Q$ and $\neg P\lor Q$ are equivalent, i.e., have the same truth value under any assignment of truth values to $P$ and $Q$. Thus, $P\to Q\iff \neg P\lor Q$ is a statement (theorem) in our natural language. In contrast, $(P\to Q)\leftrightarrow(\neg P\lor Q)$ is not a statement (claim, theorem); it is a formula. It is a tautology, i.e., it is true under all assignments of truth values to $P$ and $Q$.

This is a little like the distinction between relations like $=$, $<$ and $\le$ and operations like $+$, $-$ and $/$ in arithmetic. An expression that contains only numbers and $+$, $-$, $\cdot$ and $/$ has a numerical value, but it is nether true nor false. When we connect two such expressions by a relation such as $=$, we get a statement (claim, theorem) in our natural language. It does not make sense to write $(1=2)+3$ or $(1=1)=(2<3)$. If $E_1,E_2,E_3$ are numerical expressions, then $E_1=E_2<E_3$ is a contraction that means $E_1=E_2$ and $E_2<E_3$ and not something like $(E_1=E_2)<E_3$. Similarly, it makes sense to write $P\implies Q\implies R$ to mean $P\implies Q$ and $Q\implies R$, but this is different from $P\to Q\to R$, which is simply a formula equivalent to $P\land Q\to R$ (since usually $P\to Q\to R$ is parsed as $P\to (Q\to R)$).

I am writing this because my using $\iff$ in $V\in V\iff V\notin V$ may have led you to think that $V\in V$ and $V\notin V$ are both true. I meant that $V\in V\leftrightarrow V\notin V$ is true, which, as I said, is a contradiction because a formula $P\leftrightarrow\neg P$ is false for any truth value of $P$.

 

[evinda]

Correct.No, the last part should be $\varphi(w,w,A)$ because $w$ is not bound (quantified) in the formula. It's like when you have a formula that has 0 and $x$ and you instantiate $x$ to 0: you don't rename the existing 0.Again, the last $A$ is not replaced by some unknown $D$. $A$ is a parameter in the original formula, so it should be treated like a constant. The end of the resulting formula should be $C\in A\cup\{A\}\Rightarrow C\in A$.

Any proposition of the form $P\leftrightarrow\neg P$ is a contradiction. Whether $P$ is true or false, the whole proposition is false.

A ok... (Nod)

We have, for example, $P\to Q\iff \neg P\lor Q$. This means that the formulas $P\to Q$ and $\neg P\lor Q$ are equivalent, i.e., have the same truth value under any assignment of truth values to $P$ and $Q$. Thus, $P\to Q\iff \neg P\lor Q$ is a statement (theorem) in our natural language. In contrast, $(P\to Q)\leftrightarrow(\neg P\lor Q)$ is not a statement (claim, theorem); it is a formula. It is a tautology, i.e., it is true under all assignments of truth values to $P$ and $Q$.

Are the formulas $P\to Q$ and $\neg P\lor Q$ equivalent, since:

if $P$ is true, from $P\to Q$, $Q$ is also true and so $\neg P\lor Q = F \lor T=T$

and if $P$ is false, this: $P\to Q$ is always true and so: $\neg P\lor Q=T \lor T {/} F=T$

? Or am I wrong? (Thinking)

Similarly, it makes sense to write $P\implies Q\implies R$ to mean $P\implies Q$ and $Q\implies R$, but this is different from $P\to Q\to R$, which is simply a formula equivalent to $P\land Q\to R$ (since usually $P\to Q\to R$ is parsed as $P\to (Q\to R)$).

Why is this: $P\to Q\to R$ equivalent to $P\land Q\to R$? (Thinking)

I am writing this because my using $\iff$ in $V\in V\iff V\notin V$ may have led you to think that $V\in V$ and $V\notin V$ are both true. I meant that $V\in V\leftrightarrow V\notin V$ is true, which, as I said, is a contradiction because a formula $P\leftrightarrow\neg P$ is false for any truth value of $P$.

So, when we have $P \Leftrightarrow Q$, does this mean that both $P$ and $Q$ have to be true, but if we say that $P \leftrightarrow Q$ does this mean that $P,Q$ are both true or both false? or have I understand it wrong? (Thinking)

 

[Evgeny.Makarov]

Are the formulas $P\to Q$ and $\neg P\lor Q$ equivalent, since:

if $P$ is true, from $P\to Q$, $Q$ is also true and so $\neg P\lor Q = F \lor T=T$

and if $P$ is false, this: $P\to Q$ is always true and so: $\neg P\lor Q=T \lor T {/} F=T$

?

Strictly speaking, what you showed is that if $P=F$, then both $P\to Q$ and $\neg P\lor Q$ are true, and if $P=T$ and $(P\to Q)=T$, then $(\neg P\lor Q)=T$. What remains is that if $P=T$ and $(\neg P\lor Q)=T$, then $(P\to Q)=T$.

The easiest way to show that two formulas are equivalent is to construct their truth tables and show that they have the same values for all possible values of variables.

Why is this: $P\to Q\to R$ equivalent to $P\land Q\to R$?

You can construct truth tables for $P\to (Q\to R)$ and $P\land Q\to R$. Alternatively, consider when both formulas are false. This happens only for one set of values of $P,Q,R$.

So, when we have $P \Leftrightarrow Q$, does this mean that both $P$ and $Q$ have to be true

No. It means that $P$ and $Q$ have the same value for all possible values of variables occurring in $P$ and $Q$. In the arithmetic analogy, $P$ and $Q$ are expressions containing numbers and variables and $P=Q$ for all values of variables. For example, $(x+y)^2=x^2+2xy+y^2$. It does not means that $P$ and $Q$ always have some fixed value, but that they are always equal. Similarly, in logic $P\iff Q$ does not mean that $P=Q=T$, but that the truth values of $P$ and $Q$ are equal regardless of the values of variables.