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3. Let X be the set of positive integers. For each ne X, let Sn {k € X: k2n}. Show that T = {Sn: ne X} U {0} is a topology for X, and find the closure of the set of even integers. Find the closure of the singleton set A = {100}.

3. Let X be the set of positive integers. For each ne X, let Sn {k € X: k2n}. Show that T = {Sn: ne X} U {0} is a topology for X, and find the closure of the set of even integers. Find the closure of the singleton set A = {100}.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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3. Let X be the set of positive integers. For each n E X, let Sn {k € X: k≥n}. Show that T = {Sn: ne X} U {0} is a topology for X, and find the closure of the set of even integers. Find the closure of the singleton set A = {100}.
Transcribed Image Text:3. Let X be the set of positive integers. For each n E X, let Sn {k € X: k≥n}. Show that T = {Sn: ne X} U {0} is a topology for X, and find the closure of the set of even integers. Find the closure of the singleton set A = {100}.
Def Let (XT) be a top. space. If SEX, T-)tmit then pех на pt. of Sif meets S very in a el. of T cont. p P pt. Other than ex X=IR, T= {0, Ⓡ Suppole SER and SØ. Show cI(S)= IR. Ple+ (pEX=R). The only el. of is IR and T cont. Thm 2.2 let (X₁T) be a top. Space. If SEX [PECIS) every el. of I cont. p then meets S Thm 23 let (X, TJ be a top. space. If SCX, Sis T-closed S'ES then ef→ Assume (SEX is T-closed) Show S'ES. Notice S'ES'US = CISS Assume (S'ES Show S=CIS "C" SESUSECI (5) "2" cls=SUSSUS = S P IR meets S. By Thm 2.2, PECI(S). SoX=R)=cl S. Show c1(S) ≤ X=IR. CI S=SUS ER UR = IR Then clS= R Claim R is closed in (X,T) ef Since IR SIR and R #ø, we get cI (IR) = IR ₁ So R is closed. Claim if SER is a proper subset of IR, then S is NOT closed.
Transcribed Image Text:Def Let (XT) be a top. space. If SEX, T-)tmit then pех на pt. of Sif meets S very in a el. of T cont. p P pt. Other than ex X=IR, T= {0, Ⓡ Suppole SER and SØ. Show cI(S)= IR. Ple+ (pEX=R). The only el. of is IR and T cont. Thm 2.2 let (X₁T) be a top. Space. If SEX [PECIS) every el. of I cont. p then meets S Thm 23 let (X, TJ be a top. space. If SCX, Sis T-closed S'ES then ef→ Assume (SEX is T-closed) Show S'ES. Notice S'ES'US = CISS Assume (S'ES Show S=CIS "C" SESUSECI (5) "2" cls=SUSSUS = S P IR meets S. By Thm 2.2, PECI(S). SoX=R)=cl S. Show c1(S) ≤ X=IR. CI S=SUS ER UR = IR Then clS= R Claim R is closed in (X,T) ef Since IR SIR and R #ø, we get cI (IR) = IR ₁ So R is closed. Claim if SER is a proper subset of IR, then S is NOT closed.
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