# Defining things in logical statements

1. Nov 26, 2013

### 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.

2. Nov 26, 2013

### 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 $\forall X, \forall \mathcal S \neg\left[\Phi(X, \mathcal S)\wedge \neg \Psi(X, \mathcal S)\right]$.

3. Nov 26, 2013

### 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 $B$?

4. Nov 28, 2013

### Stephen Tashi

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.

5. Nov 28, 2013

### 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?