Question about NBG

[X.Y. Newberry]

Suppose a, b are classes, a is not a set. Suppose further that I say:

a e b

Is this syntactically incorrect?

--
X.Y. Newberry

If Jack says ‘What I am saying at this very moment is not true’, we can successfully and truly assert that he did not utter a truth: ‘What Jack said is not true’. But it is hardly conceivable that Jack’s utterance is true by virtue of its success in attributing non-truth to itself.

Haim Gaifman

[Jim Burns]

[Re: Question about NBG]

On 8/28/2014 5:25 PM, X.Y. Newberry wrote:
> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
> a e b
>
> Is this syntactically incorrect?
>

It's been a while, but I think not.
As I remember it, not being an element of anything is
what would make a a proper class,
so SET(a) <-> (E b)(a e b)

[Ross A. Finlayson]

What's the class of all classes?

It's the group noun game,
to the Inconsistency of Kunen,
or past, as it were, here,
with no winning strategy,
just overtime forever.

[Jim Burns]

[Re: Question about NBG]

If I remember correctly, there is no
class of all classes in NBG, just a
proper class of all sets.

[Mike Terry]

"Jim Burns" <burn...@osu.edu> wrote in message
news:53FFBDFA...@osu.edu...

..which is how I recall things working too. So syntactically, e applies to
classes (of which sets are a special case), and

a e b

is *syntactically* OK, but would always be false given a is not a set...
But maybe someone who works with NBG will confirm!

Mike.

[William Elliot]

On Thu, 28 Aug 2014, X.Y. Newberry wrote:

> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
> a e b
>
> Is this syntactically incorrect?

No. It a syntactically correct, false statement.

[graham...@gmail.com]

Unlike yours I assert!

[Rosario193]

On Fri, 29 Aug 2014 01:05:52 -0700, William Elliot <ma...@panix.com>
wrote:

could be true if that is followed from axioms

[Peter Percival]

If it does then NBG is indeed No Bloody Good.

[X.Y. Newberry]

So "the adjective 'heterological' is heterological" is false?

> But maybe someone who works with NBG will confirm!
>
> Mike.
>
>

--
X.Y. Newberry

If Jack says �What I am saying at this very moment is not true�, we can
successfully and truly assert that he did not utter a truth: �What Jack
said is not true�. But it is hardly conceivable that Jack�s utterance is

[Peter Percival]

X.Y. Newberry wrote:
> Mike Terry wrote:
>> "Jim Burns" <burn...@osu.edu> wrote in message
>> news:53FFBDFA...@osu.edu...
>>> [Re: Question about NBG]
>>>
>>> On 8/28/2014 5:25 PM, X.Y. Newberry wrote:
>>>> Suppose a, b are classes, a is not a set. Suppose further that I say:
>>>>
>>>> a e b
>>>>
>>>> Is this syntactically incorrect?
>>>>
>>>
>>> It's been a while, but I think not.
>>> As I remember it, not being an element of anything is
>>> what would make a a proper class,
>>> so SET(a) <-> (E b)(a e b)
>>
>> ..which is how I recall things working too. So syntactically, e
>> applies to
>> classes (of which sets are a special case), and
>>
>> a e b
>>
>> is *syntactically* OK, but would always be false given a is not a set...
>
> So "the adjective 'heterological' is heterological" is false?

You may need to explain the connection between that and NBG!

>> But maybe someone who works with NBG will confirm!
>>
>> Mike.
>>
>>
>
>


--

[Dancing is] a perpendicular expression of a horizontal desire.
G.B. Shaw quoted in /New Statesman/, 23 March 1962

[X.Y. Newberry]

Peter Percival wrote:
> X.Y. Newberry wrote:
>> Mike Terry wrote:
>>> "Jim Burns" <burn...@osu.edu> wrote in message
>>> news:53FFBDFA...@osu.edu...
>>>> [Re: Question about NBG]
>>>>
>>>> On 8/28/2014 5:25 PM, X.Y. Newberry wrote:
>>>>> Suppose a, b are classes, a is not a set. Suppose further that I say:
>>>>>
>>>>> a e b
>>>>>
>>>>> Is this syntactically incorrect?
>>>>>
>>>>
>>>> It's been a while, but I think not.
>>>> As I remember it, not being an element of anything is
>>>> what would make a a proper class,
>>>> so SET(a) <-> (E b)(a e b)
>>>
>>> ..which is how I recall things working too. So syntactically, e
>>> applies to
>>> classes (of which sets are a special case), and
>>>
>>> a e b
>>>
>>> is *syntactically* OK, but would always be false given a is not a set...
>>
>> So "the adjective 'heterological' is heterological" is false?
>
> You may need to explain the connection between that and NBG!

I thought it was obvious. First of all the class of heterological
adjectives cannot be a set because it is an equivalent of Russel's set

R = {x| x ~e x)

So let h be the proper class of heterological adjectives. We now ask if
the adjective 'heterological' is heterological. In terms of set theory
we ask is

h e h

We know that h is not the set so the above cannot be true, he he he. So
what is it then? If it is false then the adjective heterological is not
heterological.

>
>>> But maybe someone who works with NBG will confirm!
>>>
>>> Mike.
>>>
>>>
>>
>>
>
>


--

[Herman Rubin]

On 2014-08-28, X.Y. Newberry <newbe...@gmail.com> wrote:
> Suppose a, b are classes, a is not a set. Suppose further that I say:

> a e b

> Is this syntactically incorrect?

It is syntactically correct, as syntax does not know what the
variables are allowed to be. Logically, it is incorrect, as
a class is a set in NBG iff it is an element of anything.


--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

[X.Y. Newberry]

Herman Rubin wrote:
> On 2014-08-28, X.Y. Newberry <newbe...@gmail.com> wrote:
>> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
>> a e b
>
>> Is this syntactically incorrect?
>
> It is syntactically correct, as syntax does not know what the
> variables are allowed to be. Logically, it is incorrect, as
> a class is a set in NBG iff it is an element of anything.

So "the adjective 'heterological' is heterological" is false?

--
X.Y. Newberry

If Jack says �What I am saying at this very moment is not true�, we can
successfully and truly assert that he did not utter a truth: �What Jack
said is not true�. But it is hardly conceivable that Jack�s utterance is

[Herman Rubin]

On 2014-08-31, X.Y. Newberry <newbe...@gmail.com> wrote:
> Herman Rubin wrote:
>> On 2014-08-28, X.Y. Newberry <newbe...@gmail.com> wrote:
>>> Suppose a, b are classes, a is not a set. Suppose further that I say:

>>> a e b

>>> Is this syntactically incorrect?

>> It is syntactically correct, as syntax does not know what the
>> variables are allowed to be. Logically, it is incorrect, as
>> a class is a set in NBG iff it is an element of anything.

> So "the adjective 'heterological' is heterological" is false?

More precisely, it is not defined. The adjectives for color are
not defined in abstract entities, for example. Is "green"
heterological? It does not make sense.

[X.Y. Newberry]

Herman Rubin wrote:
> On 2014-08-31, X.Y. Newberry <newbe...@gmail.com> wrote:
>> Herman Rubin wrote:
>>> On 2014-08-28, X.Y. Newberry <newbe...@gmail.com> wrote:
>>>> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
>>>> a e b
>
>>>> Is this syntactically incorrect?
>
>>> It is syntactically correct, as syntax does not know what the
>>> variables are allowed to be. Logically, it is incorrect, as
>>> a class is a set in NBG iff it is an element of anything.
>
>> So "the adjective 'heterological' is heterological" is false?
>
> More precisely, it is not defined. The adjectives for color are
> not defined in abstract entities, for example. Is "green"
> heterological? It does not make sense.

It is defined: An adjective is heterological iff it does not have the
property it expresses.

We can prove that the class of heterological adjectives is NOT a set:
Assume 'heterological' is a member of the class of heterological
adjectives ... etc. A contradiction follows. Hence the class of
heterological adjectives is not a set. It *CANNOT* be an element of any
class. But what exactly does "cannot" mean?

What does it mean "defined in abstract entities"? "Green" is not
heterological because it is not green. What does not make sense? And if
"it" (whatever that is) does not make sense what are we supposed to
conclude?

a) '~(h e h)' is false
b) '~(h e h)' is malformed
c) other

[Aatu Koskensilta]

"X.Y. Newberry" <newbe...@gmail.com> writes:

> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
> a e b
>
> Is this syntactically incorrect?

Immensely incorrect, of course, and also completely correct,
naturally. It all depends on the details of the formalization of NBG you
have in mind. In some circumstances NBG is most conveniently thought of
as a first-order theory -- so that "a in b" is perfectly well-formed,
and provably false when a and b are classes -- sometimes it's best to
take NBG to be a two-sorted first-order theory -- so that whether "a in
b" is a grammatical gaffe depends on what sort of variables a and b are
-- on occasion taking NBG to be second-order ZFC in a system of
predicative second-order logic feels just right, and so on. Why?

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[Aatu Koskensilta]

Herman Rubin <hru...@skew.stat.purdue.edu> writes:

> On 2014-08-28, X.Y. Newberry <newbe...@gmail.com> wrote:
>> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
>> a e b
>
>> Is this syntactically incorrect?
>
> It is syntactically correct, as syntax does not know what the
> variables are allowed to be.

Sometimes syntax knows, sometimes it doesn't. When we formalize NBG as
a two-sorted first-order theory, for instance, the syntax turns out to
be, by construction, well aware of the distinction between set and class
variables. Such details and formalities are rarely of any mathematical
interest or consequence.

[Aatu Koskensilta]

"X.Y. Newberry" <newbe...@gmail.com> writes:

> So "the adjective 'heterological' is heterological" is false?

Now there's a baffler for the ages, to be sure. But what does it have
to do with tedious syntactic technical details in this or that formal
treatment of NBG?

[X.Y. Newberry]

Aatu Koskensilta wrote:
> "X.Y. Newberry" <newbe...@gmail.com> writes:
>
>> So "the adjective 'heterological' is heterological" is false?
>
> Now there's a baffler for the ages, to be sure. But what does it have
> to do with tedious syntactic technical details in this or that formal
> treatment of NBG?

The class of heterological adjectives is analogical to the class R of
all classes that are not members of themselves. It does not make much
sense to say that

"The adjective 'heterological' is heterological" is false

because then it is not heterological, in which case it is heterological.
Similarly

"R e R" is false

implies that ~(R e R). How do we prevent the conclusion that R e R?

[Martin Shobe]

On 9/12/2014 10:21 PM, X.Y. Newberry wrote:
> Aatu Koskensilta wrote:
>> "X.Y. Newberry" <newbe...@gmail.com> writes:
>>
>>> So "the adjective 'heterological' is heterological" is false?
>>
>> Now there's a baffler for the ages, to be sure. But what does it have
>> to do with tedious syntactic technical details in this or that formal
>> treatment of NBG?
>
> The class of heterological adjectives is analogical to the class R of
> all classes that are not members of themselves. It does not make much
> sense to say that
>
> "The adjective 'heterological' is heterological" is false
>
> because then it is not heterological, in which case it is heterological.
> Similarly
>
> "R e R" is false
>
> implies that ~(R e R). How do we prevent the conclusion that R e R?
>

What makes you think there is a class of all classes that are not
members of themselves?

Martin Shobe

[Rosario193]

above for what i think there is nothing wrong...

in my theory for mathematic "ER ReR" is a false proposition
because it is against one axiom i follow:

�(BeB)

[i put that as axiom for a reason... but not remember here...]

so �(ER ReR) is followed from axioms
and it is true for that theory
for definition of true

[X.Y. Newberry]

Because unrestricted comprehension applies to classes. But in any case
there is a class of all sets that are not members of themselves.

> Martin Shobe

[X.Y. Newberry]

Rosario193 wrote:
> On Sat, 13 Sep 2014 11:47:52 -0500, Martin Shobe
> <martin...@yahoo.com> wrote:
>
>> On 9/12/2014 10:21 PM, X.Y. Newberry wrote:
>>> Aatu Koskensilta wrote:
>>>> "X.Y. Newberry" <newbe...@gmail.com> writes:
>>>>
>>>>> So "the adjective 'heterological' is heterological" is false?
>>>>
>>>> Now there's a baffler for the ages, to be sure. But what does it have
>>>> to do with tedious syntactic technical details in this or that formal
>>>> treatment of NBG?
>>>
>>> The class of heterological adjectives is analogical to the class R of
>>> all classes that are not members of themselves. It does not make much
>>> sense to say that
>>>
>>> "The adjective 'heterological' is heterological" is false
>>>
>>> because then it is not heterological, in which case it is heterological.
>>> Similarly
>>>
>>> "R e R" is false
>>>
>>> implies that ~(R e R). How do we prevent the conclusion that R e R?
>>>
>>
>> What makes you think there is a class of all classes that are not
>> members of themselves?
>>
>> Martin Shobe
>
> above for what i think there is nothing wrong...
>
> in my theory for mathematic "ER ReR" is a false proposition
> because it is against one axiom i follow:
>

> �(BeB)

>
> [i put that as axiom for a reason... but not remember here...]
>

> so �(ER ReR) is followed from axioms

> and it is true for that theory
> for definition of true

Right. The class R of all classes that are not members of themselves is
not a set hence it is not a member of itself.

I guess I am asking how we avoid a contradiction if the non-membership
is NOT a syntactical issue.

[Rosario193]

On Sun, 14 Sep 2014 18:54:27 -0700, "X.Y. Newberry"
<newbe...@gmail.com> wrote:

>Rosario193 wrote:
>> On Sat, 13 Sep 2014 11:47:52 -0500, Martin Shobe
>> <martin...@yahoo.com> wrote:
>>
>>> On 9/12/2014 10:21 PM, X.Y. Newberry wrote:
>>>> Aatu Koskensilta wrote:
>>>>> "X.Y. Newberry" <newbe...@gmail.com> writes:
>>>>>

>> in my theory for mathematic "ER ReR" is a false proposition
>> because it is against one axiom i follow:
>>

>> �(BeB)

>>
>> [i put that as axiom for a reason... but not remember here...]
>>

>> so �(ER ReR) is followed from axioms

>> and it is true for that theory
>> for definition of true
>
>Right. The class R of all classes that are not members of themselves is
>not a set hence it is not a member of itself.
>
>I guess I am asking how we avoid a contradiction if the non-membership
>is NOT a syntactical issue.

all mathematic is about how to write formulas followed from axioms

it is all "a syntactical issue"
it is about the right combination of symbols follow some laws

[Martin Shobe]

On 9/14/2014 8:49 PM, X.Y. Newberry wrote:
> Martin Shobe wrote:
>> On 9/12/2014 10:21 PM, X.Y. Newberry wrote:
>>> Aatu Koskensilta wrote:
>>>> "X.Y. Newberry" <newbe...@gmail.com> writes:
>>>>
>>>>> So "the adjective 'heterological' is heterological" is false?
>>>>
>>>> Now there's a baffler for the ages, to be sure. But what does it
>>>> have
>>>> to do with tedious syntactic technical details in this or that formal
>>>> treatment of NBG?
>>>
>>> The class of heterological adjectives is analogical to the class R of
>>> all classes that are not members of themselves. It does not make much
>>> sense to say that
>>>
>>> "The adjective 'heterological' is heterological" is false
>>>
>>> because then it is not heterological, in which case it is heterological.
>>> Similarly
>>>
>>> "R e R" is false
>>>
>>> implies that ~(R e R). How do we prevent the conclusion that R e R?
>>>
>>
>> What makes you think there is a class of all classes that are not
>> members of themselves?
>
> Because unrestricted comprehension applies to classes. But in any case
> there is a class of all sets that are not members of themselves.

Class comprehension in NGB doesn't allow quantifiers over classes. In
any case, NGB is well-founded, so no set is a member of itself.
Therefore, the class of all sets that are not members of themselves is
simply the class of all sets. There's still no analogue of heterological
here.

Martin Shobe

[X.Y. Newberry]

Are quantifications over classes syntactically incorrect or false?

> In
> any case, NGB is well-founded, so no set is a member of itself.
> Therefore, the class of all sets that are not members of themselves is
> simply the class of all sets.

OK, so let H be the class of all sets. Is

H e H

syntactically incorrect or false? What about

~(H e H) ?

> There's still no analogue of heterological
> here.

> Martin Shobe
>

[Rosario193]

it is easy
H, the set of all set
not exist
in mathematic...

all the proposition speak of H speak out math theory...

[Peter Percival]

It's false. The class of all sets is a proper class and therefore not
an element of anything.

> What about
>
> ~(H e H) ?

True because the underlying logic is classical and thus two-valued.

>> There's still no analogue of heterological
>> here.
>
>> Martin Shobe
>>
>
>


--

[Peter Percival]

Rosario193 wrote:
> On Wed, 17 Sep 2014 18:48:48 -0700, "X.Y. Newberry"

>>

>> OK, so let H be the class of all sets. Is
>>
>> H e H
>>
>> syntactically incorrect or false? What about
>>
>> ~(H e H) ?
>>
>>> There's still no analogue of heterological
>>> here.
>>
>>> Martin Shobe
>>>
>
> it is easy
> H, the set of all set
> not exist
> in mathematic...

Mr N. is not talking about the set of all sets, but rather the *proper*
class of all sets. Hence the "NBG" in his Subject header. See, for
example,
http://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory.

[Martin Shobe]

The syntax is fine. They just won't be instances of class comprehension.

>> In
>> any case, NGB is well-founded, so no set is a member of itself.
>> Therefore, the class of all sets that are not members of themselves is
>> simply the class of all sets.

> OK, so let H be the class of all sets. Is
>
> H e H

> syntactically incorrect or false? What about

It's false.

> ~(H e H) ?

It's true.

Martin Shobe

[Shmuel Metz]

In <lvef1m$188$1...@news.albasani.net>, on 09/18/2014

at 12:18 PM, Peter Percival <peterxp...@hotmail.com> said:

>True because the underlying logic is classical and thus two-valued.

No. It's true because it follows from the axioms and the rules of
inference. The underlying logic of NF is classical and two valued, but
~(H e H) is false.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[X.Y. Newberry]

So why can I not say that there is no adjective "heterological" only a
meta-adjective "heterological". Meta-adjectives have the property that
no adjective or meta-adjective can be applied to them. Then
"heterological" is not heterological because no adjective or
meta-adjective can be applied to it.

[Martin Shobe]

> So why can I not say that there is no adjective "heterological" only a
> meta-adjective "heterological". Meta-adjectives have the property that
> no adjective or meta-adjective can be applied to them. Then
> "heterological" is not heterological because no adjective or
> meta-adjective can be applied to it.

It probably wouldn't match the grammar rules for any current natural
language (at least I'm not aware of any that it matches), but you
certainly could do that for some language you are constructing.

Now what does that have to do with NGB?

Martin Shobe

[X.Y. Newberry]

NBG permits "heterological" in a meta-language. It basically has one
layer of meta-language. The meta-language can speak about the language,
but the meta-language cannot speak about itself. That is, there are
heterological adjectives in the language. But we cannot speak about
"heterological" because it is in the meta-lalanguage.

In ZFC there is no adjective "heterological". "Heterological" is
ill-defined and there is no such concept.

This brings me to the following question. We know there is no Russell
set R. ~(ER)(R = {x | ~(x e x)}). But what if I write

R e R ?

Is is syntactically malformed or is it false? How about

~(R e R)
?

[Martin Shobe]

I don't see the connection you are trying to make between
"Heterological" and NGB (or ZFC). Maybe you should try making what part
of NGB (and ZFC) corresponds to "Heterological" explicit.

> This brings me to the following question. We know there is no Russell
> set R. ~(ER)(R = {x | ~(x e x)}). But what if I write

> R e R ?

> Is is syntactically malformed or is it false? How about

It's false. (I'm assuming you are talking about ZFC now and not
repeating your questions about NGB).

> ~(R e R)

[X.Y. Newberry]

Is it not obvious that the concept of "heterological" is analogical to
the set of all sets that are not members of themselves?

>> This brings me to the following question. We know there is no Russell
>> set R. ~(ER)(R = {x | ~(x e x)}). But what if I write
>
>> R e R ?
>
>> Is is syntactically malformed or is it false? How about
>
> It's false. (I'm assuming you are talking about ZFC now and not
> repeating your questions about NGB).
>
>> ~(R e R)
>> ?
>
> It's true.
>
> Martin Shobe
>

[Martin Shobe]

No. Especially since the first time I asked, you said it was the class
of all classes that are not members of themselves. Still, I suppose you
could look at it that way. I still don't see what the point of it would be.

[snip]

Martin Shobe

[graham...@gmail.com]

On Monday, September 1, Newberry wrote:
>
> It is defined: An adjective is heterological iff it does not have the
>
> property it expresses.

ok so

H = { long , ... }

short ~e H

since 'short' is a short adjective.

>
>
>
> We can prove that the class of heterological adjectives is NOT a set:
>
> Assume 'heterological' is a member of the class of heterological
>
> adjectives ... etc. A contradiction follows. Hence the class of
>
> heterological adjectives is not a set. It *CANNOT* be an element of any
>
> class. But what exactly does "cannot" mean?

stratified out, unable to be specified, re-defined

>
>
>
> What does it mean "defined in abstract entities"? "Green" is not
>
> heterological because it is not green. What does not make sense? And if
>
> "it" (whatever that is) does not make sense what are we supposed to
>
> conclude?
>
>
>
> a) '~(h e h)' is false
>
> b) '~(h e h)' is malformed
>
> c) other
>
>
>
>

> Right. The class R of all classes that are not members of themselves is
> not a set hence it is not a member of itself.
>
> I guess I am asking how we avoid a contradiction if the non-membership
> is NOT a syntactical issue.

You could use a more advanced grammar than WFF.

http://tinyurl.com/new-math-foundations


Well Formed Formula are a syntactic method to
recursively incrementally construct finite sentences.

WFF ::= term_n
WFF ::= VAR_n
WFF ::= term_n( WFF_1 , WFF_2 , ... WFF_n )
WFF ::= ALL( VAR_n ) WFF
WFF ::= EXIST( VAR_n ) WFF
WFF ::= WFF_1 ^ WFF_2
WFF ::= WFF_1 v WFF_2
WFF ::= WFF_1 -> WFF_2
WFF ::= WFF_1 <-> WFF_2
WFF ::= WFF_1 = WFF_2
WFF :: = ~WFF

e.g. by limiting ALL(V) WFF
to ALL(V) V<epsilon

so you can range over ANY ELEMENT in formulations
but not ALL ELEMENTS





Or you could try my new Logic Quantifier (V)

http://tinyurl.com/new-logic-foundations

RUSSELL SET 1
Ax xer <-> x~ex

RUSSELL SET 2
r = Ax xer <-> (x~ex & x~=r)

RUSSELL SET 3
Vx xer <-> x~ex




RS2 & RS3 are equivalent and valid sets

"the set of all sets that don't belong themselves barring this actual set!"

RS3 is a concise notation that avoids Russells Paradox without any restrictions on naïve set specification.




Just define the set correctly to start with!

heterological = 'properties that are not properties of that property barring heterological itself'

DONE! DEFINED!

[graham...@gmail.com]

> > heterological adjectives is not a set. It *CANNOT* be an element of any
> >
> > class. But what exactly does "cannot" mean?
>
>
>
> stratified out, unable to be specified, re-defined
>

I might add: backtracking to a resolved model

but this is O(n^2) as there are

[VXV] comparisons to find a contradiction.



www.tinyurl.com/new-logic-foundations

[George Greene]

On Thursday, August 28, 2014 5:25:14 PM UTC-4, Newberry wrote:
> Suppose a, b are classes, a is not a set. Suppose further that I say:
>
>
>
> a e b
>
>
>
> Is this syntactically incorrect?

In the usual syntax, no. But people can innovate in ways that would make it incorrect. Usually, however, a class-theory implemented in a single-sorted first-order language (which is the "usual" default) will allow this, syntactically. It will by definition be false if a is not a set, but it
is syntactically well-formed and therefore gets to have a truth-value.