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I need help with the exercise 6. page 14, in the notes "Simply Connected Spaces" by Dr. John Lee (*Notes-Total number of pages are 14). The problem says that if we take an open and path connected subset U of $\mathbb R^2$ and we delete countably infinite points $x_1,x_2,... \in$ U, then the set U - {$x_1,x_2,...$} is multiply connected. (or not?)

*** Edited: We note that "Multiply connected topological space is a space which is path-connected but not simply connected."

I solved the first question in this exercise, which was to show that the set U - {$x_1,x_2,...$} is path connected, by using the fact that infinite lines pass through a point in U. But I do not know any machinery from algebraic topology to use the fundamental group here. I have also found a source that is similar to my problem, in a question here: $\pi_{1}({\mathbb R}^{2} - {\mathbb Q}^{2})$ is uncountable , but I could not still solve this completely.

Another idea I had for a solution:

First I solved the exercise Exercise 4. page 14 in Dr Lee's Notes:
"Suppose A is any path connected subset of $\mathbb R^2$ that contains some circle but not it's center. Prove that A is multiply connected."
~ (Quick sketch of my proof: I solved this by finding a retraction map from A to the circle, which we know that is multiply connected, because any circle is homeomorphic with the circle $ \mathbb S^1 $, which we have shown that is multiply connected. The construction of the retraction map - formula is similar with the formula of the retraction map we found to prove that the punctured plane $\mathbb R^2$- {(0,0)} is multiply connected.)

Second, we can use the result of Dr Lee's - Exercise 4, here in this problem, so we can have a very quick solution, not only for the case of deleting countably infinite points of U, but also for the case of deleting a finite set of points in U. It is easy to find circles which are contained in U - {$x_1,x_2,...$} and their centers are not contained in U - {$x_1,x_2,...$}, using the fact that U is open set in $\mathbb R^2$, we can find open disks with centers the $x_1,x_2,...$ and radius for example $\varepsilon $>0. So if we fix, for example, a radius $\varepsilon $ /2, we will take the circles we need, which are contained in U - {$x_1,x_2,...$} and their centers $x_1,x_2,...$ are not contained in U - {$x_1,x_2,...$}. But I do not know if my solution is correct, it seems ok, but it seems too easy in this way for both cases, so i am afraid I made a terrible mistake and I do not have someone to verify my answers.

I tried many days to solve this before I ask here, I do not have a professor to help me, it is self-study Topology for deeper understanding in Manifold Theory which I love the most. Thank you for any response, I would appreciate any help!

Spyridwn Angelopoulos
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Your solution is pretty much correct. You can complete it by asking yourself why it doesn’t work if we remove an uncountably infinite number of points. You didn’t consider whether the circle is actually contained in $U-\{x_1,x_2,\ldots\}$; in fact it might not be if it contains one of the excised points $x_1,x_2,\ldots$. However, since uncountably infinitely many values of $\epsilon$ are available and only countably many of the corresponding circles can contain one of the excised points, you can choose $\epsilon$ such that the circle is in fact contained in $U-\{x_1,x_2,\ldots\}$.

joriki
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    Thank you very much for the response! I understand completely now thanks to your answer, it was in front of me but I could not see! Bless you :) – Spyridwn Angelopoulos Feb 26 '22 at 19:24
  • Well, I used your way and it was very good, but I think I found another way too, without using the "uncountably or countably many" - argument. It can be applied for both cases, for finite and countably infinite points in U. I chose the radius of every circle to be in such way so the diacenter (a.k.a. the line which connects the centers of two circles) of one circle to every other circle, to be "bigger" than the sum of the radiuses of both circles. So all the circles will be "far away" each other, informally speaking. Is it right for both cases ? We do not need to count circles in this way... – Spyridwn Angelopoulos Feb 27 '22 at 02:44
  • @SpyridwnAngelopoulos: It's not clear to me how you make that choice – can you describe it more formally? – joriki Feb 27 '22 at 15:17
  • We can find open disks $D(x_i,R_i),\forall\ i=1,...$ with centers the excised points $x_i$ and radiuses $R_i.$ Then we fix radiuses $r_i$ for the same centers $x_i$ which satisfy the conditions: $0<r_i<R_i$ and the distance $d(x_i,x_j)>r_i+r_j$ , $\forall\ i\neq j,\ \forall\ i,j=1,...$
    Let's note the new circles as $S(x_i,r_i), \forall\ i=1,...$
    They will be externally disjoint each other. So any other excised point - center $x_j \notin S(x_i,r_i),\ \forall\ i\neq j,\ \forall\ i,j=1,...$ So we do not need to count circles as a part of our solution and we are calm for both cases!     – Spyridwn Angelopoulos Feb 27 '22 at 16:51
  • @SpyridwnAngelopoulos: I'm not sure about the part "Then we fix radiuses $r_i$ for the same centers $x_i$ which satisfy the conditions...". One of the points could be an accumulation point of the remaining points, and then any circle around it, no matter how small we make it, would contain other points. – joriki Feb 27 '22 at 18:44
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    You have right! I did not think about that, so it can hold only for the "finite" case. Thank you very much for the help and the kindness! – Spyridwn Angelopoulos Feb 27 '22 at 21:32