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let d be the euclidean metric of $\mathbb R^2$

let

W={$(x,y):x^2+y^2<4$} $\mathbb U$ {$(x,y):x^2+y^2$ $<4 $ and $x^2+y^2\leq9$}

X={$(x,x):x\in \mathbb R$} $\mathbb U$ {$(x,-x):x\in \mathbb R$}

Y={$(q,1):q\in \mathbb Q$}

Z={$(x,y):(x+2)^2+y^2=4$} $\mathbb U$ {$(x,y):(x-2)^2+y^2<4$}

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 tempted to say the set $S_{1}$={$(x,y):x^2+y^2<4$} is open as it is just an open subset of the reals and the set $S_2$= {$(x,y):x^2+y^2$ $\leq9 $}-{ $x^2+y^2<4$} is open in W because although the edge of the circle of radius of 9 is included in the set, i can put "balls" around these points as there are no points in W inside my "balls that go outside the edge of the circle. So W=$S_1$$\mathbb U$$S_2$ and this is a disjoint union. is this anywhere near correct?

i have the same trouble for the other sets. i'm guessing

X is connected as its just a cross

Y is not connected as i have gaps in the line through y=1 at each irrational

Z is connected as it is 2 circles that touch at the origin and the origin is included in the space Z

which leads me how do i show it is connected? do i have to show there are no open subsets of X whose union is the whole of X?

jiboom
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    Use \cup ($\cup$) instead of \mathbb{U} for union – Matthew Leingang Apr 08 '18 at 23:09
  • Saul : doesn't that inequality give the boundary of W? your comment on X suggests it is connected, so how do I know it has no open subsets? – jiboom Apr 09 '18 at 11:18
  • Saul: can i say that any subset of X is going to be union of points (x,x) or (x,-x) but if then to say these are open i require a set S in $\mathbb R^2$ such that {(x,x)}=X n S. but such an S must be the point (x,x) along with any of $\mathbb R^2$ not containing X and this cannot be open as i cant put a ball around (x,x) as i have no other points on the line contained in X – jiboom Apr 09 '18 at 20:55
  • @Saulspatz : this refers to my previous comments. just seen I needed to do @ before your name. – jiboom Apr 10 '18 at 09:58
  • @Mike A : after a search you seem to have followed same course. did you study this module? can you offer any suggestions ? – jiboom Apr 10 '18 at 10:00

1 Answers1

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Does this work for W?

i have W={$(x,y):x^2+y^2<4$} $\mathbb U$ {$(x,y):x^2+y^2$ $<4 $ and $x^2+y^2\leq9$}

so could i let

S={$(x,y):x^2+y^2<4$}

T={$(x,y):4<x^2+y^2\leq 9$}

so W=S$\cup$T and S and T are disjoint

now

S = W $\mathbb n$ {$(x,y):x^2+y^2<4$}

and {$(x,y):x^2+y^2<4$} is open in $\mathbb R^2$

and can i say

T=W$\mathbb n${$(x,y):4<x^2+y^2<10$}

and as $\mathbb n${$(x,y):4<x^2+y^2<10$} is open in $\mathbb R^2$ then

T is open in W

the conclusion is then since i can write W as the union of two disjoint W open sets it is therefore disconnected?

im still unsure about the other sets, if someone could even just confirm i have the correct connected/disconnected for X ,Y, Z so i know im in the right area to try and proceed.

thanks

jiboom
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  • Isn't $\mathbf W$ supposed to be ${(x,y)|x^2+y^2<4}\cup {(x,y)|4<x^2+y^2<9}?$ To say $x^2+y^2<4\text{ and } x^2+y^2<9$ is the same as to say $x^2+y^2<4$. – saulspatz Apr 10 '18 at 12:57
  • @Saulspatz , yes I have the inequality around the wrong way. W is just the circle of radius 9 with the circumference of the circle of radius 4 removed. would my answer about suffice? – jiboom Apr 10 '18 at 14:21
  • Yes, it's correct. – saulspatz Apr 10 '18 at 14:25
  • @Saulspatz , yes I have the inequality around the wrong way. W is just the circle of radius 9 with the circumference of the circle of radius 4 removed. would my answer about suffice? for X can I say any subset of it is union of points,and points are not open so only subsets that are open and closed are empty set and X? this just leaves Z. I can't,now,even decide if it's connected. I thought at first it is because (0,0) is included in the set but I also have circumference of a circle so can I write this as intersection of Z with a closed/open set in $\mathbb R^2$ – jiboom Apr 10 '18 at 14:29
  • @Saulspatz how do we know the line has no open subsets? in another question I have to find an open cover of the set $P={(x,sin x):x\in \mathbb R} $ with respect to the induced metric on P (induced from euclidean metric). does this not curve not have non open subsets similar to the line case? – jiboom Apr 16 '18 at 21:36