/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Prove: (a) \(\left(S_{1} \cap ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 13

Prove: (a) \(\left(S_{1} \cap S_{2}\right)^{0}=S_{1}^{0} \cap S_{2}^{0}\) (b) \(S_{1}^{0} \cup S_{2}^{0} \subset\left(S_{1} \cup S_{2}\right)^{0}\)

Short Answer

Expert verified
In conclusion, we have proven the two set operations involving the closure of sets. We showed that the closure of the intersection of two sets is equal to the intersection of their closures, \((S_1 \cap S_2)^0 = S_1^0 \cap S_2^0\), and that the union of the closures of two sets is a proper subset of the closure of their union, \(S_1^0 \cup S_2^0 \subset (S_1 \cup S_2)^0\). These properties of set closures support a better understanding of how set operations interact with the concept of closure and can be helpful in different fields of mathematics, such as topology and analysis.

Step by step solution

01

Part A: Proving \((S_1 \cap S_2)^0 = S_1^0 \cap S_2^0\)#

To prove the equality, we will show that \((S_1 \cap S_2)^0 \subseteq S_1^0 \cap S_2^0\) and \(S_1^0 \cap S_2^0 \subseteq (S_1 \cap S_2)^0\).
02

Proving \((S_1 \cap S_2)^0 \subseteq S_1^0 \cap S_2^0\)#

Let \(x \in (S_1 \cap S_2)^0\). Then, \(x\) is either an element of \(S_1 \cap S_2\) or a limit point of \(S_1 \cap S_2\). If \(x \in S_1 \cap S_2\), then \(x \in S_1\) and \(x \in S_2\). So, \(x \in S_1^0\) and \(x \in S_2^0\), since the closure contains all elements of the set. Thus, \(x \in S_1^0 \cap S_2^0\). If \(x\) is a limit point of \(S_1 \cap S_2\), then for every open neighborhood \(U_x\) of \(x\), \(U_x\) contains an element of \(S_1 \cap S_2\) distinct from \(x\). Since \(S_1 \cap S_2 \subseteq S_1\) and \(S_1 \cap S_2 \subseteq S_2\), \(U_x\) will also contain elements of \(S_1\) and \(S_2\) distinct from \(x\). Therefore, \(x\) is a limit point of both \(S_1\) and \(S_2\). Hence, \(x \in S_1^0 \cap S_2^0\). Thus, \((S_1 \cap S_2)^0 \subseteq S_1^0 \cap S_2^0\).
03

Proving \(S_1^0 \cap S_2^0 \subseteq (S_1 \cap S_2)^0\)#

Let \(x \in S_1^0 \cap S_2^0\). Then, \(x\) is either an element or a limit point of both \(S_1\) and \(S_2\). If \(x\) is an element of both \(S_1\) and \(S_2\), then \(x \in S_1 \cap S_2\). So, \(x \in (S_1 \cap S_2)^0\), since the closure contains all elements of the set. If \(x\) is a limit point of both \(S_1\) and \(S_2\), then for every open neighborhood \(U_x\) of \(x\), \(U_x\) contains an element of both \(S_1\) and \(S_2\) distinct from \(x\). Therefore, \(U_x\) contains an element of \(S_1 \cap S_2\) distinct from \(x\). Hence, \(x\) is a limit point of \(S_1 \cap S_2\). So, \(x \in (S_1 \cap S_2)^0\). Thus, \(S_1^0 \cap S_2^0 \subseteq (S_1 \cap S_2)^0\). Since both inclusions are true, we can conclude that \((S_1 \cap S_2)^0 = S_1^0 \cap S_2^0\).
04

Part B: Proving \(S_1^0 \cup S_2^0 \subset (S_1 \cup S_2)^0\)#

To prove this, we will show that every element of \(S_1^0 \cup S_2^0\) is also an element of \((S_1 \cup S_2)^0\).
05

Proving \(S_1^0 \cup S_2^0 \subseteq (S_1 \cup S_2)^0\)#

Let \(x \in S_1^0 \cup S_2^0\). Then, either \(x \in S_1^0\) or \(x \in S_2^0\). This means that \(x\) is either an element or a limit point of either \(S_1\) or \(S_2\). If \(x\) is an element of \(S_1\), then \(x \in S_1 \cup S_2\). Therefore, \(x \in (S_1 \cup S_2)^0\), since the closure contains all elements of the set. Similarly, if \(x\) is an element of \(S_2\), then \(x \in (S_1 \cup S_2)^0\). If \(x\) is a limit point of \(S_1\), then for every open neighborhood \(U_x\) of \(x\), \(U_x\) contains an element of \(S_1\) distinct from \(x\). Consequently, \(U_x\) also contains an element of the set \(S_1 \cup S_2\) distinct from \(x\). Therefore, \(x\) is a limit point of \(S_1 \cup S_2\) and \(x \in (S_1 \cup S_2)^0\). Similarly, if \(x\) is a limit point of \(S_2\), then \(x \in (S_1 \cup S_2)^0\). Thus, \(S_1^0 \cup S_2^0 \subseteq (S_1 \cup S_2)^0\). Since it is given that it is a proper subset, we have proved that \(S_1^0 \cup S_2^0 \subset (S_1 \cup S_2)^0\).

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