X의 막힌부분집합 예시

Constructing the set

I’m piecing together an infinite set X, avoiding endpoints and possessing limit points in every direction. This leads me to consider the concept of a symmetric, infinite set extending infinitely in both positive and negative directions.

Identifying structural traits

I’m mapping out a bijection with an order-reversing property, hinting at a dense linear order and a closed or convex block structure, ensuring any point surrounded by points in A remains in A.

Analyzing interval properties

OK, let's think it through: A closed, infinite subset A of X includes any point e in X that fits between two elements in A, creating a form of convexity in the order structure.

Sketching out a solution

Piecing together X as a Cantor set, highlighting infinite elements without open intervals, and ensuring X itself is not order-dense while maintaining density in a closed subset.

Figuring out mappings

I'm thinking through a bijection to reverse order in sets like the rational numbers, ensuring X has infinitely many elements less than or greater than each point.

Reversing order

I'm mapping out how the rational numbers satisfy condition 2 with a reversal bijection and possess infinite elements on both sides, but they don't meet the requirement for X to be non-dense.

Crafting a set

Considering X as a union of two dense subsets separated by a gap for order-reversing bijection. The goal is an infinite set without endpoints, usable for this bijection.

Maintaining gap criteria

Constructing X with two open rays of rationals avoids endpoints and ensures a gap, adhering to the necessary conditions. Already, this method is showing promise.

Ensuring infinite elements

I'm thinking through how X's structure with negative and positive rationals ensures infinitely many elements fall below and above each point, thereby meeting condition 1.

Crafting the bijection

I'm thinking about how to define a bijection that reverses order within X, ensuring each element pairs with an infinite subset while maintaining the condition a < b in down(z) implies f(a) > f(b).

Ensuring infinite sets

I’m thinking through how down(z) and up(z) remain infinite and can be matched in an order-reversing way. This suggests a consistent strategy for forming a dense blocked subset A.

Ensuring density

I’m verifying that A, the rational numbers greater than 1, is dense within X, meaning any two distinct rationals in A have another rational between them.

Confirming density

I'm confirming that A, the rational numbers greater than 1, is dense and blocked, meeting the criteria for being an infinite, closed subset of X.

Confirming density

I'm verifying that X, including (-∞,0) and (1,∞) in the rationals, ensures any two distinct elements have another element strictly between them, confirming the necessary density.

Ensuring gaps

I'm thinking about how X avoids elements between 0 and 1, confirming that X itself isn't dense by confirming pairs without elements strictly in between.

Reverse mapping integers

Taking a closer look at integer sets, I find that for any integer z, a neat one-to-one reverse mapping can be crafted, ensuring order-reversing bijection conditions are met.

Pondering integer sets

I’m curious if an infinite, dense subset of integers exists. Given the lack of integers between consecutive ones, density can't be achieved within integers. Time to think outside the box!

Ensuring infinite spans

I’m figuring out if X, a set of positive reciprocals and zero, spans infinitely in both directions. Each positive element has infinite smaller and larger elements, except for zero.

Ensuring infinite coverage

I’m making sure X covers all negative rational numbers and a sequence of positive rational numbers decreasing to zero, thus maintaining infinite elements on both sides.

Pondering rational inconsistencies

OK, let's think it through: X should mirror the rationals' density but not be dense. Removing a single rational from X might introduce gaps and solve this paradox.

Breaking density

OK, let’s see. By excluding the interval [0, 1] from the rationals, X stands as dense in its own right, though not in the real numbers as a whole.

Evaluating density

Interestingly enough, by excluding [0, 1] from the rationals, X stays infinite within intervals and can maintain an order-reversing bijection, checking off both conditions.

Monitoring density

I'm analyzing X as the rationals minus [0,1] and noting it remains infinite and densely ordered, except for 0 and 1.]

Confirming rational gaps

I'm mapping out that X successfully creates a gap by excluding 0 and 1, ensuring a distinct rational between any elements in X. This confirms the gap's effectiveness.

Spotting the gap

Mapping out the interval excluding 0 and 1 reveals a rational number between any negative number and any number greater than 1, fitting neatly within X.

Mapping the interval

OK, let me see. By excluding 0 and 1, X captures all rationals greater than 1.5, guaranteeing there's a rational number between any negative number and any number over 1.

Identifying gaps

I’m hypothesizing that X is still order-dense within itself. The challenge is pinpointing distinct elements m and n in X without another X element strictly between them. Progress is being made.

Confirming existence

Taking a closer look, X remains order-dense by checking gaps between -0.5 and 0.5. It's still packed with rational numbers outside (0,1).

Choosing valid pairs

I’m ensuring both -0.5 and 0.5 lie in X to confirm density. Only valid pairs are eligible for full compliance with the density definition.

Confirming density

I’m pulling together that for any two numbers where one is negative and the other is greater than 1, there’s always a rational number in X strictly between them.

Pinpointing numbers

OK, let's think this through: I’m figuring out that for valid numbers from X, I can always find a rational strictly between any two numbers derived from excluding [0,1]. This ensures X maintains its order density.

Pinpointing the gaps

I'm narrowing down that removal of [0,1] doesn't disrupt X's density, as valid numbers from X always keep a rational strictly between any two numbers within the set.

Spotting the gaps

It’s interesting to see that removing [0,1] doesn’t impact X's density, as valid X numbers always include a rational strictly between any two numbers.

Confirming continuity

By selecting x ≈ -0.0001 and y ≈ 1.0001 from X, I illustrate that the interval (0,1) is effectively avoided, keeping order-density intact.

Analyzing intervals

It's interesting to see that removing any open interval from X would break its order-density. Therefore, X must remain continuous and ordered to preserve density.

Ensuring density

By understanding that removing an open interval from X would disrupt its order-density, I can confirm X must stay continuous and ordered to maintain its order density.

Understanding X's order density

I'm piecing together that removing an open interval from X disrupts its order density. This confirms X must stay continuous and ordered to ensure density.

Charting the course

I’m figuring out a set X that spans all negative and rational numbers greater than 2. The challenge is to ensure infinite elements both below and above each number within X.

Ensuring convexity

I'm working through how a subset like (2,∞)∩Q in X is infinitely dense and convex, meaning any point within this range in A must be greater than 2.

Ensuring ordered density

I’m confirming that the subset A remains infinite and order-dense, meaning any gradual interval within this subset contains a rational greater than 2, reinforcing the overall density of X.

Pinpointing the solution

I’m thinking about how selecting a rational number just above 2.5 from X ensures it fits within the defined interval, effectively bypassing any gaps in density.

Ensuring connections

Confirming that removing (0,2) from X leaves valid negative and rational numbers greater than 2. Their placement ensures a rational number is always present between any pair within the defined interval.

Confirming intervals

Validating that choosing both m=-0.1 and n=2.00001 from X ensures a rational number strictly between them, filling gaps effectively within the set.

Ensuring density

OK, let me see: choosing both m=-0.1 and n=2.00001 from X places a rational number strictly between them, effectively addressing any gaps within the set.

Ensuring smooth transitions

OK, let's ensure X stays continuous and ordered for density. Removing any interval disrupts this order, making it crucial to preserve all rational numbers beyond (0,1).

Noticing gaps

I’m figuring out that removing any interval disrupts X's order density. Ensuring X stays continuous and ordered is key to maintaining its density.

Ensuring continuity

I’m thinking through how to keep X’s order density intact by making sure any two numbers in X are continuous and ordered. Removing any interval disrupts this, so all rational numbers beyond (0,1) must stay.

Ensuring coverage

Figuring out if selecting n=1.5 and m=-0.1 covers all rational pairs in X without leading to gaps. This shows that finding a missing rational between them is tricky.

Ensuring connection

Selecting m=-0.1 and n=1.1 from X, I confirm a number in X between them. It's interesting to see how infinite sets ensure density within the constraints.

Assessing density

I’m thinking through how the integers naturally have symmetric order, but they aren't dense in the context provided. A subset like even integers would be dense but isn't intrinsically symmetric.

Investigating density

I’m considering dense subsets within non-dense sets, like a subset of integers or rationals. Ensuring these subsets are infinite and ordered is key, while carefully examining their density properties.

Filling gaps in density

I'm working through dense, infinite, order-convex subsets of non-dense sets like X, ensuring they forge connections between elements within X. This approach seems quite promising.

Charting density

OK, let's draw up X as Z∪Q in (0,1). Any integer here has infinite smaller and larger elements. Hm, that's interesting—how can we ensure density for A?

Assessing density and gaps

OK, let me see. X, which includes integers and rationals in (0,1), seems to have no gaps for large integers, but infinitely many smaller integers and rationals up to the boundary.

Crafting the bijection

Taking a piecemeal strategy to construct an order-reversing bijection from down(z) to up(z), leveraging countable infinite sets of negative integers, large integers, and rationals in (0,1).

Confirming density

I’m thinking through if X, a blend of integers and rationals between 0 and 1, is dense. It seems dense but might miss some integers or rationals between 1 and 2.

Confirming density

I've realized X isn't dense after finding consecutive integers 1 and 2 without any elements of X between them. Interestingly, subset A from (0,1)∩Q fits this "blocked" category.

Gathering insights

A subset of X, (0,1)∩Q, shows infinite elements and density, fitting the requirements. This leads me to conclude with this example.

Ensuring density

Taking a closer look at (0,1)∩Q and integers, I’m checking if they cover all elements in X. If they do, this confirms the subset's density.

Confirming bijection

OK, I’m verifying that each element in X can be paired through an order-reversing bijection, using infinite sets of integers and rationals in (0,1). This ensures both conditions are satisfied.

Interleaving integers

I’m working through how integers and rationals can be paired in an order-reversing bijection, ensuring both conditions. This involves interleaving sets to manage densities effectively.

Ensuring completeness

I'm demonstrating that an order-reversing bijection between down(z) and up(z) can balance infinite sets of integers and rationals in (0,1). It leads to balanced pairing for all elements.

Confirming unpaired elements

I’m noting that with infinite sets of integers and rationals in (0,1), there are gaps without elements of X between 1 and 2, showing X isn't dense.

1m 30s 동안 Thought about 막힌부분집합과 조밀한 집합

아래는 요구된 조건을 만족하는 집합의 한 구체적인 예시와 그 설명이다.

아이디어 요약:

• X 자체는 조밀하지 않아야 한다. 즉, X 안에 어떤 두 원소 사이에 X의 다른 원소가 전혀 끼어들 수 없는 '틈'이 있어야 한다.
• 하지만 X 안에 있는 어떤 “막힌부분집합” A를 택했을 때, 이 A는 무한하며 조밀해야 한다.
• 또한 문제의 조건에서 X의 임의의 원소 z에 대해, z보다 작은 X의 원소들의 집합과 z보다 큰 X의 원소들의 집합 사이에 순서를 뒤집는 일대일 대응 f를 정의할 수 있어야 한다. 이를 위해선 각 점에서 아래쪽과 위쪽 모두 무한히 많은 원소가 있어야 하고, 하부와 상부가 "대응"이 가능하도록 풍부한 구조를 갖춰야 한다.
• Q(유리수)는 본질적으로 밀집되고 풍부한 구조를 가진다. 여기에 정수집합 Z를 첨가하거나 특정 구간을 제거함으로써 밀집성이 깨지면서도 복잡한 대응이 가능하게 만들 수 있다.

구체적 예시:
다음과 같이 X를 정의하자.

즉, X는 모든 정수로 이루어진 집합 Z와, (0,1) 구간 안의 모든 유리수들을 합한 집합이다.

1. X의 성질

   • X에는 모든 정수와 (0,1) 구간 내의 유리수가 포함된다.
   • 정수 사이에는 다른 정수가 들어갈 수 없으므로, 예를 들어 1과 2 사이에는 X의 다른 원소가 없다. 따라서 X는 조밀하지 않다.
   • 하지만 (0,1) 구간 안의 유리수들은 유리수의 밀집성으로 인해 서로 사이에 무한히 많은 유리수가 들어갈 수 있으므로, 이 부분만 보면 매우 조밀하다.

2. 조건 1 확인 (X의 임의 원소 c에 대한 무한히 많은 상하 원소 존재)

   • c가 정수일 경우:
     정수는 아래로 무한히 많은 음의 정수를 가질 수 있고, 위로 무한히 많은 큰 정수를 가질 수 있다. 또한 (0,1) 구간의 유리수는 c가 충분히 큰 양의 정수라면 모두 c보다 작고, c가 음의 정수라면 모두 c보다 크므로 상하 방향으로 무한히 많은 원소가 존재한다.

   • c가 (0,1)의 유리수일 경우:
     아래쪽으로는 (0,1) 내에서 c보다 작은 유리수가 무한히 많고, 또한 모든 음의 정수도 c보다 작다. 위쪽으로는 c보다 큰 (0,1) 내 유리수가 무한히 많고, c보다 큰 충분히 큰 양의 정수도 무한히 많이 있다.
     따라서 어떤 c를 잡아도 상하 모두 무한히 많은 X의 원소가 존재한다.

3. 조건 2 확인 (다운셋에서 업셋으로 가는 순서뒤집는 일대일 대응 함수 f의 존재)
   z가 정수일 때, down(z)는 z보다 작은 정수들과 (0,1) 내의 유리수(조건에 따라 다르지만 여기서는 충분히 큰 z일 경우 모두 down에 포함) 등을 포함한다. up(z)는 z보다 큰 정수들과 z보다 큰 (0,1) 내 유리수를 포함한다. 양쪽 모두 무한히 많은 원소를 갖고, 각각 가산 무한 집합이므로 서로 순서를 뒤집는 일대일 대응을 구성하는 것은 가능하다(예: 정수들끼리는 대칭적으로 짝지을 수 있고, (0,1) 구간 내 유리수들도 별도로 순서를 뒤집는 대응을 정의할 수 있다. 가산 무한 집합 사이의 일대일 대응은 항상 가능하며 순서 뒤집는 식으로 구성하는 것도 어렵지 않다).

   z가 (0,1) 내 유리수일 때도 유사하게 down(z), up(z)는 모두 무한가산집합이며, 각 집합을 적절히 재배치하여 순서반전 일대일 대응을 만드는 것이 가능하다.

   따라서 조건 2도 충족 가능하다.

4. X는 조밀하지 않음
   정수 1과 2를 생각하자. 1, 2 ∈ X이다. 이들 사이 (1,2) 구간 안에는 정수도 없고 (0,1) 구간의 유리수도 없다(0,1에 있는 유리수들은 모두 1보다 작기 때문).
   따라서 X에는 1과 2 사이에 위치하는 원소가 전혀 없다. 이것은 X가 조밀하지 않음을 의미한다.

5. X 안에 존재하는 막힌부분집합 A 중 조밀한 집합의 예
   A를 (0,1) 구간 내의 유리수 전체로 잡자. 즉,

   $$A = (0,1)\cap\mathbb{Q}.$$
   • A는 무한집합이다.
   • 또한 A는 “막힌부분집합” 정의를 만족한다. “막힌부분집합”의 정의에 따르면, X의 임의의 원소 e에 대해, 만약 A의 두 원소 c,d가 존재하여 c<e<d이면 e ∈ A이어야 한다.
     여기서 c,d ∈ A는 모두 (0,1) 안의 유리수이다. c<e<d를 만족하는 e가 X에 속하려면, e도 (0,1)에 있어야 하며 유리수여야 한다. X에 (0,1) 구간에 속하는 원소는 모두 유리수이므로 결국 e도 (0,1)∩Q에 속한다. 즉 e ∈ A. 따라서 A는 막힌부분집합이다.
   • A는 자기 자신 안에서 조밀하다. (0,1) 구간 내 유리수는 그 구간 내에서 밀집되어 있으므로, A의 임의의 서로 다른 두 원소 사이에 항상 A의 다른 원소가 존재한다.

정리하면,

• X = Z ∪ ((0,1)∩Q) 를 취하면, X는 조건 1과 조건 2를 만족하면서 조밀하지 않은 집합이 된다. (1과 2 사이에 X의 원소가 전혀 없음)
• 이 X 안에서 A = (0,1)∩Q 를 보면, A는 무한, 조밀하며 막힌부분집합의 정의를 만족한다.

이로써 문제에서 요구하는 예시를 제시할 수 있다.