如何证明两类集合族的特定关系？含集合论等式证明任务

First, let's clear up your confusion about the definition of $U$—it's not a subset of $F$, which seems to be the root of your confusion. Let's rephrase $U$ in plain terms to make it concrete:

$U = {X \subseteq A : \text{every set } S \text{ in } F \text{ is entirely contained within } X}$

In other words, $U$ is the collection of all supersets of every element in $F$. For example, if $A = {1,2,3,4}$ and $F = {{1,2}, {2,3}}$, then $U$ would include sets like ${1,2,3}$ and ${1,2,3,4}$—any subset of $A$ that contains both elements of $F$. It's abstract at first, but it's just the family of all "common supersets" for $F$.

Proving $\bigcup F = \bigcap U$ #

We'll use the standard set equality technique: show each set is a subset of the other.

Step 1: $\bigcup F \subseteq \bigcap U$ #

Take any element $x \in \bigcup F$. By definition of union, there exists some $S_0 \in F$ where $x \in S_0$. Now, pick any $X \in U$—by $U$'s definition, $S_0 \subseteq X$, so $x$ must be in $X$. Since this holds for every $X$ in $U$, $x$ is in the intersection of all $U$'s elements. Thus, $\bigcup F \subseteq \bigcap U$.

Step 2: $\bigcap U \subseteq \bigcup F$ #

Suppose $x \in \bigcap U$. Let's use proof by contradiction: assume $x \notin \bigcup F$, meaning $x$ isn't in any set in $F$. Consider $X_0 = A \setminus {x}$—does $X_0$ belong to $U$? For every $S \in F$, $S$ doesn't contain $x$, so $S \subseteq X_0$. That means $X_0 \in U$, but $x \notin X_0$, which contradicts $x$ being in every element of $U$. Our assumption is wrong—$x$ must be in $\bigcup F$. Thus, $\bigcap U \subseteq \bigcup F$.

Combining both steps, $\bigcup F = \bigcap U$.

Proving the Dual: $\bigcap F = \bigcup L$ for some $L \subseteq \mathcal{P}(A)$ #

This is the mirror result, using subsets instead of supersets. First, define $L$ appropriately:

$L = {Y \subseteq A : \text{every set } S \text{ in } F \text{ contains } Y}$

In short, $L$ is all subsets that are contained in every element of $F$. Now let's prove the equality.

Step 1: $\bigcup L \subseteq \bigcap F$ #

Take any $x \in \bigcup L$. There exists some $Y_0 \in L$ where $x \in Y_0$. By $L$'s definition, $Y_0$ is a subset of every $S \in F$, so $x$ is in every $S \in F$. Thus, $x \in \bigcap F$. So $\bigcup L \subseteq \bigcap F$.

Step 2: $\bigcap F \subseteq \bigcup L$ #

Take any $x \in \bigcap F$. Consider the singleton set $Y_x = {x}$. Is $Y_x$ in $L$? Since $x$ is in every $S \in F$, ${x}$ is a subset of every $S \in F$, so $Y_x \in L$. And since $x \in Y_x$, $x$ is in $\bigcup L$. Thus, $\bigcap F \subseteq \bigcup L$.

Combining both steps, $\bigcap F = \bigcup L$.

Key Takeaway #

The core idea here is a fundamental duality in set theory: unions can be represented as intersections of all their supersets, and intersections can be represented as unions of all their subsets. Your initial mix-up about $U$ being a subset of $F$ is common—remember, $U$ is the family of sets that contain every element of $F$, not a subset of $F$ itself.

内容的提问来源于stack exchange，提问作者oldselflearner1959