Defining things in logical statements

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V0ODO0CH1LD
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Is the following a valid logical proposition?

Proposition: If ##\mathcal{S}## is a collection of subsets of ##X## that covers ##X## then the collection ##\mathcal{B}## of all finite intersections of elements of ##\mathcal{S}## is a basis for a topology on ##X##.

Firstly, when I state a proposition, is it implied that what I mean is "proposition: it is true that if ... then ..."? if not, what does it mean to propose an implication?

Anyway, I ask if the above is a valid logical proposition because the second statement in that implication is not defined on its own, it needs the first statement since ##\mathcal{S}## is defined as a collection of subsets of ##X## only on the first statement. Would ##\mathcal{S}## qualify as a free variable? In which case the proposition is not a logical proposition in classical logic?

I guess my question could be generalized to: in classical logic, if I propose that ##x\rightarrow y##, is it a convention that what I am proposing is that ##x\rightarrow y=1##, and also, could I make that proposition if ##y## only makes sense if ##x## is true?

NOTE: The topology example is only an example to help me illustrate my question, it is not the actual question.
 
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You have two formulas, which take [itex]X[/itex] and [itex]\mathcal S[/itex] as parameters.
- [itex]\Phi(X,\mathcal S)[/itex], which says that [itex]\mathcal S \subseteq 2^X[/itex] and [itex]\bigcup\mathcal S = X[/itex].
- [itex]\Psi(X,\mathcal S)[/itex], which says that [itex]\left\{\bigcap\mathcal F: \enspace \mathcal F \subseteq\mathcal S \text{ is finite} \right\}[/itex] is a basis for a topology on [itex]X[/itex].

The proposition says "[itex]\Phi(X, \mathcal S)\implies \Psi(X, \mathcal S)[/itex]", which (as you noted) reads as a formula with free variables [itex](X, \mathcal S)[/itex]. That said, it's very common that when people write the proposition [tex]\Phi(X, \mathcal S)\implies \Psi(X, \mathcal S)[/tex] (with the variables having no specific prior meaning), what they really mean is [tex]\forall X, \forall \mathcal S \left[\Phi(X, \mathcal S)\implies \Psi(X, \mathcal S)\right][/tex] which is in turn a shorthand for [itex]\forall X, \forall \mathcal S \neg\left[\Phi(X, \mathcal S)\wedge \neg \Psi(X, \mathcal S)\right][/itex].
 
I think I don't know what a free variable in logic is then. I though a free variable in a logic statement was not allowed since it could render the statement "unknown". I still feel, even in your optimized way of stating that proposition, that in ## \Psi(X,\mathcal{S}) ## the variable ## S ## is not even defined, so ##\mathcal F \subseteq\mathcal S## makes no sense at all.

Another question; to prove this proposition, the route taken is usually to assume that ## \Phi(X,\mathcal{S}) ## is true, and given that show that ## \Psi(X,\mathcal{S}) ## is also true. Although an implication is still true if the first statement is false. My question is, when in a math textbook the author writes "proposition: ## A\implies B ##", does he mean "##A=1##, proposition: ## A\implies B=1 ##"?

And also, is it okay to define a variable in ##A## and then use it in ##B##?
 
V0ODO0CH1LD said:
I think I don't know what a free variable in logic is then. I though a free variable in a logic statement was not allowed since it could render the statement "unknown".

Common mathematical writing and correct logical statements are two different things. If someone says "if x is greater than 0 then 2x is greater than 0", he has technically not said whether he means to modify x by "for each number x" or "there exists a number x", but the normal "cultural" interpretation of such a statement is that the quantifier "for each number x" is to be applied to x.
 
Okay, let me ask a few question to see if I can clarify this in my head.

A proposal, like a theorem, can proven or disproven.

Strictly speaking, proposing what is called a formula in logic doesn't make any sense. Like "I propose that x + y = 1", makes no sense as a proposal because it's true for some values of x and y and untrue for others. But "I propose that for all x there exists a y such that x + y = 1", is a valid proposal since it can be reduce to true or false. Right?