# connected spaces/topology/metric spaces

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Thread starter 3 years ago
#1
let d be the euclidean metric of let W= and X= Y= Z= I have to decide if the above sets are d-connected or d-disconnected. i'm inclined to say W is not connected as its the circle of radius 9 take away the circumference of circle radius 4. so for example i cant get to the origin from a point near edge of big circle.

But to prove this i need to write W as union of 2 disjoint sets that are d-closed and d-open.( to do this according to my book i need to say, as **W** is subset of the reals any closed subset of **W** is intersect of **W** and **F** where **F** is a closed set in $\mathbb R^2$ hopefully ive understood the theory correctly,but now i'm confused how to find my subsets.

im thinking i need to write W as union of the 2 sets and and is open since its the intersection of W and and is open in and is open since its the intersection of W and and is open in does this look close?

any help with the others will be appreciated
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3 years ago
#2
I'm not clear why you think you need to do this. The normal definition would be W is disconnected if you can find two non-empty open sets that partition W, and it's obvious that and work for this (your argument seems OK, but I confess I'd be happier with an epsilon-proof here).

[These sets are also closed (relative to W), but I don't see why you actually need to show this].
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Thread starter 3 years ago
#3
(Original post by DFranklin)
I'm not clear why you think you need to do this. The normal definition would be W is disconnected if you can find two non-empty open sets that partition W, and it's obvious that and work for this (your argument seems OK, but I confess I'd be happier with an epsilon-proof here).

[These sets are also closed (relative to W), but I don't see why you actually need to show this].
(Original post by DFranklin)
I'm not clear why you think you need to do this. The normal definition would be W is disconnected if you can find two non-empty open sets that partition W, and it's obvious that and work for this (your argument seems OK, but I confess I'd be happier with an epsilon-proof here).

[These sets are also closed (relative to W), but I don't see why you actually need to show this].
thank you for the reply. wasnt sure you still posted here,but i recall you having a flair for this terrible analysis stuff!! firstly the definitions i have to work with are

S is disconnected if S is union of 2 disjoint clopen sets

although it does say later on we only need the sets to be open or closed.

connected if only 2 subsets that are clopen are the empty set and the whole set.

for a set T in S to be closed in S i need to write T=SnR for some open set in R^2

these are the definitions im working with

i have done W now.

for Y i translated it to the x-axis and looked at the real numbers with gaps at each irrational.think this is ok

for X i think any subset of X wiil be a union of points of the lines y=x and y=-x. now these points are not open so the only 2 subsets of X that are closed and open are the empty set and X. when i say points are not open in X im really not sure of my proof.

for Z,not sure how to proceed. think its connected. if so how do i show this?
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3 years ago
#4
(Original post by mathz)
firstly the definitions i have to work with are

S is disconnected if S is union of 2 disjoint clopen sets

although it does say later on we only need the sets to be open or closed.

connected if only 2 subsets that are clopen are the empty set and the whole set.
Here's the wikipedia definition:

https://en.wikipedia.org/wiki/Connected_space

You'll see (and this is fairly normal), that there's one "fundamental" definition, and you then show that 3 or 4 other definitions are equivalent.

So (of course), the standard approach in an exam is to choose which ever definition makes life easiest.

for X i think any subset of X wiil be a union of points of the lines y=x and y=-x. now these points are not open so the only 2 subsets of X that are closed and open are the empty set and X. when i say points are not open in X im really not sure of my proof.
This isn't convincing to me. Geometrically, X is two lines that cross. If it was two lines that didn't cross (e.g. y = x and y = x+1), it wouldn't be connected. So your argument needs to use the fact that the lines cross and I'm not seeing that.

I think you'd be best off showing each line is connected, and then showing that two connected sets with non-zero intersection are connected.

for Z,not sure how to proceed. think its connected. if so how do i show this?
I may be missing something, but what I think I'd do is use "path connected => connected" (which you may need to prove, but the proof is online and easy). Then show Z is path connected. (You may also be able to translate this to a direct proof).
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Thread starter 3 years ago
#5
(Original post by DFranklin)

This isn't convincing to me. Geometrically, X is two lines that cross. If it was two lines that didn't cross (e.g. y = x and y = x+1), it wouldn't be connected. So your argument needs to use the fact that the lines cross and I'm not seeing that.

I think you'd be best off showing each line is connected, and then showing that two connected sets with non-zero intersection are connected.
.
Yes,that makes sense of my garbled argument.

for Z i dont have path connected so could i get away with this:

let

U= V= so Z is union of U and V ,U V intersect and i know from the book U is connected and,this is the winging part, V is interior of circle with an accumulation point so is also connectd.
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3 years ago
#6
(Original post by mathz)
Yes,that makes sense of my garbled argument.

for Z i dont have path connected so could i get away with this:

let

U= V= so Z is union of U and V ,U V intersect and i know from the book U is connected and,this is the winging part, V is interior of circle with an accumulation point so is also connectd.
Isn't Z path connected? To get from a point (x,y) in U to a point in V, you go from (x, y) to (x, 0) to (0,0) which is in V, and then go round the circle.
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Thread starter 3 years ago
#7
(Original post by DFranklin)
Isn't Z path connected? To get from a point (x,y) in U to a point in V, you go from (x, y) to (x, 0) to (0,0) which is in V, and then go round the circle.
Yes, z is path connected. I typed out reply but it didnt post so seems in my rush to retype I missed out the end of that line. what it should say is I dont have path connected as something from the book. im stuck with the union of connected sets etc.
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3 years ago
#8
(Original post by mathz)
Yes, z is path connected. I typed out reply but it didnt post so seems in my rush to retype I missed out the end of that line. what it should say is I dont have path connected as something from the book. im stuck with the union of connected sets etc.
I understand. But my point was that "path connected => connected" is only a 2-3 line proof. So even if you don't "know" about path connected yet, you could still use that and the proof to guide you.
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Thread starter 3 years ago
#9
(Original post by DFranklin)
I understand. But my point was that "path connected => connected" is only a 2-3 line proof. So even if you don't "know" about path connected yet, you could still use that and the proof to guide you.
thaks for the help. i have another question about open covers. if you have time could you have a look at:

i have the set let d be the euclidean metric on i need to find :

an open cover of P with no finite subcover

an open cover of P with a finite subcover

an open cover

where is the induced metric on P

for the first im thinking i can have the union of overlapping rectangles of height 2 where each overlap leaves a strip that is purely in one rectangle. so if i took a finite subcover i would have these strips missing.

for the second i would take the cover to be the union of the 2 sets

S= T= so S T covers P and my finite cover is {S,T }

the last one is worrying me,as it nust be something to do with the curve but can i say the arcs are d_{p) -open as we just had lines were not open in the connected question?
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