# Understanding of merging of junctions of collections

If you start with a boundless set, you can have a series of nested collections which merge to a solitary factor. (ie Intersection of $\left(\large\frac{-1}{n}, \frac{1}{n}\right)$ as $n\to \infty$)

However, at no time throughout the series exists a first component with just one factor. Actually, for any kind of limited (yet boundless) $n$, the variety of factors in the period is uncountably huge. So, just how do you recognize the suggestion that this boundless junction has simply one factor? Just how do you understand it?

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2019-05-13 03:31:41
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A factor remains in the junction if it remains in every set of the series. Concentrate on that. Despite just how near $0$ a factor is (leaving out $0$ itself), there will certainly be a completely huge $n$ such that the set $(-1/n,1/n)$ does not have it, and also consequently that factor will certainly not remain in the (boundless) junction, because, once more, a factor remains in the junction iff and also it remains in every set defined by the junction.

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2019-05-17 15:11:44
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I assume the resource of the complication is this : taking the cardinality of a set feels like such a straightforward procedure that we unconsciously anticipate it to be continual. We could assume that straightforward restricting procedures like countable nested junctions need to offer convergent cardinalities. Yet the straightforward reality is that they do not ; our subconscious hunch is simply incorrect, and also this instance confirms it. Cardinality is not continual in this feeling.

This relates to the typical "mystery" of the supertask where sometimes $1-2^{-n}$ you include in a box 2 rounds phoned number $2n-1$ and also $2n$, and also remove the one phoned number $n$. Although that the cardinality of the rounds in package is constantly raising, in the restriction sometimes 1 package is vacant. Once more, the mystery emerges just if you really feel that cardinality needs to be continual. If you understand that it isn't, after that you no more see such instances as paradoxical.

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2019-05-17 15:09:23
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