Proof of a proposition involving universal quantifiers and a cover

Question

I started to learn topology recently. Solving some examples in a textbook, I found the solutions for some problems make use of a cover in a way that can be described as follows (if I correctly understood):

Consider the topological space $X$, the set $S$, and the family of sets $C$ such that $\forall A \in C: A\subseteq X$, $S\subseteq X$, and $S \subseteq \cup C$, that is, $C$ covers $S$. $P$ is a predicate. Then, the following proposition seems true. $$ (\forall A \in C: (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x)) $$

But I could not figure out how to prove that. I am not familiar with logic and don't know how to correctly handle universal quantifiers.

The last question is, are $x\in S \implies P(x)$ and $\forall x \in S : P(x)$ equivalent in this case? Any advice will be appreciated.

Answer 1

Let $P=\{x\in X:P(x)\}$. Then $\forall A\in C\,\big(x\in A\cap S\to P(x)\big)$ clearly just says that $A\cap S\subseteq P$ for each $A\in C$, and $x\in S\to P(x)$ says that $S\subseteq P$. Thus, your question boils down to asking whether $\forall A\in C\,(A\cap S\subseteq P)$ is equivalent to $S\subseteq P$.

Suppose that $A\cap S\subseteq P$ for each $A\in C$, and let $x\in S$. $C$ covers $S$ so there is an $A\in C$ such that $x\in A$. Clearly $x\in A\cap S\subseteq P$, so $x\in P$, and since $x$ was an arbitrary element of $S$, we’ve shown that $S\subseteq P$.

Now suppose that $S\subseteq P$, and let $A\in C$. Then $A\cap S\subseteq S\subseteq P$, so $A\cap S\subseteq P$. Thus, $A\cap S\subseteq P$ for all $A\in C$ if and only if $S\subseteq P$.

The answer to your last question is yes.

Answer 2

1. $\forall x \in S : P(x)$ is shorthand for $\forall x, (x \in S) \wedge P(x)$, so the second answer is "absolutely".

2. For the first, let's use that second one: \begin{align} (\forall A& \in C : (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x)) \\ (\forall A &: ((A \in C ) \wedge (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x)) & \text{applying rule from above}\\ \end{align} Now let's look at that last thing. Suppose $x \in S$. Then because $C$ is a cover, there's some element $A$ of $C$ such that $x \in A$. That means that $x \in A$ and $x \in S$, so $x \in A \cap S$, so $P(x)$ holds. That proves the left-to-right implication.

Now suppose that the right-hand side is known. Let $A$ be any element of $C$. Then $x \in A \cap S$ tells use $x \in S$, hence $P(x)$ (by our assumption). So that proves the right-to-left implication.