Question

Is [0,1] \(\cup\) (2,3] connected? Prove or disprove
I don't know how to start, please, help

Answer 1

tell me which topic is it , or space

Answer 2

metric space, advance calculus, compact set part

Answer 3

First off, let A =[0,1] \(\cup\) (2,3]
need define 2 open sets U, V st......
My question is, we can arbitrary pick U, V, right?

Answer 4

ok they are disjoint

Answer 5

show that \[A \cap \overline{B} = \overline{A}\cap B = \emptyset \]

Answer 6

@ganeshie8 what do you mean by B(bar)??

Answer 7

[0,1] ∪ (2,3] well i only looked at those ,union of disjoint , so not connected

Answer 8

compliment?

Answer 9

closure of B

Answer 10

so closure in this definition :-

closure of A is intersect of all closed sets that contain A ?

Answer 11

@ikram002p you know!! we have to follow the definition and give out the proof, not intuitively say yes or no.

Answer 12

A is concluded by cl A

Answer 13

@ganeshie8 what are A, B?

Answer 14

A = [0,1]
B = (2,3]

Answer 15

i dont know which definition you follow xD
so post urs

Answer 16

@ganeshie8 it is one set with 2 parts, not 2 sets

Answer 17

ganesh this means its not connected ,right ?
http://prntscr.com/59fi9w

Answer 18

say S = [0,1] ∪ (2,3]

A = [0,1]
B = (2,3]

Answer 19

yes @ikram002p

Answer 20

Loser what def of closure do you use? There are many. A delta ball, intersection of sets, ...

Answer 21

1 second

Answer 22

ok i wanted to confirm xD
loser then u need to show they are disjoint ,wanna continue on it or u have another def?

Answer 23

They give us A has 2 parts as we have in problem, need define U, V with that property to conclude whether A connected or not.

Answer 24

so this is A ?
[0,1] u (2,3]

Answer 25

Yes

Answer 26

U and V are closures ?

U = [0, 1]
V = [2, 3]

Answer 27

they should be , but not necessary though

Answer 28

to the problem (2,3] not [2,3]

Answer 29

ok chose U and V open

Answer 30

like this :-
V=(-1,2)
U=(2,4)

Answer 31

agree so far ?

Answer 32

then U and V = empty, right?

Answer 33

yes

Answer 34

see now the three condition would satisfy

Answer 35

and \(U\cup V=(-1,4)/\{2\}\)

Answer 36

then \(A\subset U\cup V\) the last condition satisfied

Answer 37

http://prntscr.com/59fprf

Answer 38

\(A\cap V= (2,3]\neq \emptyset\)

Answer 39

yeah :)

Answer 40

see , i still think cluser definition is more proper , and make more sense

Answer 41

so A is disconnected, righ?

Answer 42

yes !

Answer 43

woohoo.. I got it, thaaaaank you

Answer 44

np :)

Answer 45

next time say its metric on R , ok ?

Answer 46

hihihi... Yes, madam.

Answer 47

xD js madam :P

Answer 48

js ??? mean??

Answer 49

how is the def'n in textbook equivalent to the def'n using closures @ikram002p

Answer 50

just saying :)

Answer 51

its not equivalent xD

Answer 52

they have to be equivalent as they are defining the same thing : "disconnected set" right ?

Answer 53

closure definition works in any space what ever is it ...
discrete , indiscreet , left ray , standard (metric ), ..... ect

so this definition , just set it in metric only

Answer 54

that makes sense thanks

Answer 55

xD