Defining things in logical statements

[V0ODO0CH1LD]

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.

 

[economicsnerd]

You have two formulas, which take X and \mathcal S as parameters.
- \Phi(X,\mathcal S), which says that \mathcal S \subseteq 2^X and \bigcup\mathcal S = X.
- \Psi(X,\mathcal S), which says that \left\{\bigcap\mathcal F: \enspace \mathcal F \subseteq\mathcal S \text{ is finite} \right\} is a basis for a topology on X.

The proposition says "\Phi(X, \mathcal S)\implies \Psi(X, \mathcal S)", which (as you noted) reads as a formula with free variables (X, \mathcal S). That said, it's very common that when people write the proposition \Phi(X, \mathcal S)\implies \Psi(X, \mathcal S) (with the variables having no specific prior meaning), what they really mean is \forall X, \forall \mathcal S \left[\Phi(X, \mathcal S)\implies \Psi(X, \mathcal S)\right] which is in turn a shorthand for \forall X, \forall \mathcal S \neg\left[\Phi(X, \mathcal S)\wedge \neg \Psi(X, \mathcal S)\right].

 

[V0ODO0CH1LD]

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 $B$?

 

[Stephen Tashi]

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.

 

[V0ODO0CH1LD]

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?