/*! 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} Q 66. Let S be a subset of R2 or R3. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let S be a subset of R2or R3. Prove that a set S is closed if and only if ∂S⊆S.

Short Answer

Expert verified

It is proved that " the set is closed if and only if∂S⊆S.

Step by step solution

01

Step 1. Given information.  

We have given S is a subset of R2or R3.

02

 Prove the given statement.  

The objective is to prove that the set is closed if and only if ∂S⊆S

Assume an element x' such that X∈S, where Sis a closed set.
By the definition of a closed set, if Sis a closed set, then its complementSe is an open set.
This means that for element 'X' there exists such an open disk or ball D, such that D would include or extend beyond the boundary of S.
Thus, the set D∩Seor∂S∩S is non-empty.
From these statements, it is clear that ∂S⊆S

In another case, consider S is an open set.

A set is said to be open if for every element of it, there exists an open disk or ball D, such thatx∈D⊆S
This would mean that 'x' does not belong to the boundary ofS.
x∉∂S
Thus, there is no common element between Sand ∂SThis means that as ∂S∉S
Combining the two proofs, it is proved that, "that the set is closed if and only if∂S⊆S

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.