Suppose SCC and z €Cis a boundary point of S. Choose the statement which cannot be false. O a. There exists a neighbourhood of z that is completely contained in S O b. Every neighbourhood of z is completely contained in S Oc There exists a neighbourhood of z that is completely contained in the complement of S. O d. Every neighbourhood of z is completely contained in the complement of S. O e. Every deleted neighbourhood of z contains at least one point in S and at least one point not in S O L Every neighbourhood of z contains at least one point in S and at least one point not in S

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Suppose SCC and z e C is a boundary point of S. Choose the statement which cannot be false.
O a. There exists a neighbourhood of z that is completely contained in S
O b. Every neighbourhood of z is completely contained in S
O c. There exists a neighbourhood of z that is completely contained in the complement of S.
O d. Every neighbourhood of z is completely contained in the complement of S.
O e. Every deleted neighbourhood of z contains at least one point in S and at least one point not in S
f. Every neighbourhood of z contains at least one point in S and at least one point not in S
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2022-03-27
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Transcribed Image Text:W 9 - U - Document1 - Microsoft Word (Product Activation Failed) File Home Insert Page Layout References Mailings Review View % Cut A Find - Calibri (Body) - 11 - 三、三, 情 | T A A Aa Aal AaBbCcDc AaBbCcDc AaBbC AaBbCc AaB AaBbCcL E Copy Сopy a Replace B I U abe x, x A х, х* ab I Normal I No Spaci. Heading 1 Paste Change Styles - Select - Heading 2 Title Subtitle W Format Painter Clipboard Font Paragraph Styles Editing L • 2:1: 1: | 3:1' 4: ·5.1 6.1:7 l:8: 1'9 ' 10: 1 '11: 1'12 :L·13:1' 14:' 15. LA:L 17:L · 18. Suppose SCC and z e C is a boundary point of S. Choose the statement which cannot be false. O a. There exists a neighbourhood of z that is completely contained in S O b. Every neighbourhood of z is completely contained in S O c. There exists a neighbourhood of z that is completely contained in the complement of S. O d. Every neighbourhood of z is completely contained in the complement of S. O e. Every deleted neighbourhood of z contains at least one point in S and at least one point not in S f. Every neighbourhood of z contains at least one point in S and at least one point not in S ull 06:33 PM 2022-03-27 Page: 1 of 1 Words: 0 B I E E E 90% e +
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