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CK (Y,V) is cont. if =1₂5) = cl₂ ( f(s)) 1=1}₁₂ T = subspace top: from R, uval) =discrete top f(x) = { ₁ if x 4-1 if X 21 mt. et, ≤-1 and a pt. ?) case Let S S [1,00). Notice f(s) = {1}. Then clxsx = [1,00)X = [1,00 So cl(s)[, ~] Then f (cl(s)) = {1}. So f (cl 5 ) = {1} = Cly([1]) = clyl- ↑ [17 is closed in (Y, V) D Case Let S (-00,-1]. (HW)

CK (Y,V) is cont. if =1₂5) = cl₂ ( f(s)) 1=1}₁₂ T = subspace top: from R, uval) =discrete top f(x) = { ₁ if x 4-1 if X 21 mt. et, ≤-1 and a pt. ?) case Let S S [1,00). Notice f(s) = {1}. Then clxsx = [1,00)X = [1,00 So cl(s)[, ~] Then f (cl(s)) = {1}. So f (cl 5 ) = {1} = Cly([1]) = clyl- ↑ [17 is closed in (Y, V) D Case Let S (-00,-1]. (HW)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 30E
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dwu527.jpg ^ Q Search 3.1 344444443SCOGCO 66 collab.com/collab/ui/session/playback Good Morning ETS Def f: (x,T) → (Y, V) is cont. if * SEX, f(cs) ≤cl( f(s)) ex X= {x€R: 1×1² 1}¸ T= subspace top- from (iR, usual) -I claim f is Cont. Y = {0, 1} V = discrete top f: X→Y has f(x) = { 0₁ if x 4-1 if X 21 pt. bo4EEEESEEEEEO3 Case S has a We show f(clŚ) Ecl(f(s)) - 4-1 and a pt. ²1 D Q Search Δ 313 Bb X Case Let ETS MEC case Let S ≤ [1,00). Notice f(s) = {1}. Then IR clxs = c/₁RSX = [1,00)^X = [1,00), So cl(s) ≤ [1,~] 3 Then f (c1(s)) = {1}. So f (cl S) == {¹} = cly ([¹]) = cly (f(s) ) ↑ {1} is closed in (Y, V) SC (-00,-1]. (HW) SSO
Transcribed Image Text:dwu527.jpg ^ Q Search 3.1 344444443SCOGCO 66 collab.com/collab/ui/session/playback Good Morning ETS Def f: (x,T) → (Y, V) is cont. if * SEX, f(cs) ≤cl( f(s)) ex X= {x€R: 1×1² 1}¸ T= subspace top- from (iR, usual) -I claim f is Cont. Y = {0, 1} V = discrete top f: X→Y has f(x) = { 0₁ if x 4-1 if X 21 pt. bo4EEEESEEEEEO3 Case S has a We show f(clŚ) Ecl(f(s)) - 4-1 and a pt. ²1 D Q Search Δ 313 Bb X Case Let ETS MEC case Let S ≤ [1,00). Notice f(s) = {1}. Then IR clxs = c/₁RSX = [1,00)^X = [1,00), So cl(s) ≤ [1,~] 3 Then f (c1(s)) = {1}. So f (cl S) == {¹} = cly ([¹]) = cly (f(s) ) ↑ {1} is closed in (Y, V) SC (-00,-1]. (HW) SSO
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