91Ó°ÊÓ

When we talk about a family of intervals, we imagine a collection of intervals, each with their own unique bounds, grouped together under a single set. Imagine them like siblings in a real family, each interval as an individual, but related by a defining characteristic. In the context of real analysis, these families can have infinitely many members, as is the case with the set \( \mathscr{F}_{1} \) in our problem.

In our particular exercise, each interval \( I_n = \{x: 1/2^n < x < 2\} \) becomes a member of the family as \( n \) varies over the natural numbers. This family is notable for its intervals that get infinitely close to 0 as \( n \) gets larger, due to the lower bound, \( 1/2^n \), decreasing. This characteristic is crucial in covering interval \( J \), since any positive number less than 1 can be caught between the jaws of one of these intervals.

A cover of an interval is a collection of sets that together contain every point of that interval. If you can imagine painting a fence, where the sections of the fence are intervals, a cover would be enough paint buckets such that every part of the fence gets painted with no gaps left.

Using this analogy, the interval \( J = \{x: 0 < x < 1\} \) is like our fence, and we are showing that our family of intervals \( \mathscr{F}_{1} \) has enough 'paint buckets' to 'paint' the whole. To prove that \( \mathscr{F}_{1} \) is indeed a cover, we take any point \( x \) in \( J \) and argue there is at least one interval containing that \( x \). Since the family includes intervals with lower bounds that can be made arbitrarily small (but never hitting zero), it ensures that for any point in \( J \) there's an interval \( I_n \) in \( \mathscr{F}_{1} \) that 'covers' it.

When we move on to consider a finite subfamily, we are essentially choosing a limited set of intervals from our infinite family in hopes that they still do the job of covering the interval \( J \). It's like choosing a finite number of paint buckets from an infinite supply and hoping you have chosen well enough to paint the entire fence.

Unfortunately, as the steps in our solution indicate, trying to cover \( J \) with a finite subfamily of \( \mathscr{F}_{1} \) is like showing up to a mile-long fence with only a dozen buckets of paint – we're bound to run out. The trick lies in the lower bounds of the intervals in the family \( \mathscr{F}_{1} \) getting smaller as \( n \) increases. If you choose any finite number, \( N \) for example, there will always be a positive number in \( J \) that’s smaller than \( 1/2^N \) – it won't be 'painted', or in mathematical terms, not covered by any interval in our finite selection. Hence, while the entire family of intervals can cover \( J \) infinitely, no finite subset of said family can achieve the same coverage.