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anyone can help with this:

Let $X$ be the so-called Hawaiian Earrings, i.e. union of these circles:

$$\left(x − \frac1n\right)^2 + y^2 = \left(\frac1n\right)^2 , n = 1, 2, \dots\;,$$

with the induced topology of the plane and let $Y$ be the space when we identify every integer points of real line to a point. Show that $X$ and $Y$ are not homeomorphic.

Thanks

2 Answers 2

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HINT: $X$ is compact. Find a sequence in $Y$ with no convergent subsequence.

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    I'm thinking about the following sequence: 1/2, 3/2, 5/2, 7/2, ... but I don't know how to prove this sequence haven't any convergent subsequence2012-09-28
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    @user42912: That’s a good sequence to think about, because it works. Just show that every point of $Y$ has an open neighborhood that contains at most one point of this sequence. There are really only two cases: $y\in Y$ is not an integer, and $y$ is an integer. In the second case $\bigcup_{n\in\Bbb Z}\left(n-\frac12,n+\frac12\right)$ works; can you do the other case?2012-09-28
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    Let y be not an integer, just take this neighborhood (q-1/2, q+1/2), am I right?2012-09-28
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    @user42912: Not quite: what if $y=1/3$? Then you need to take a smaller neighborhood; the biggest interval that works is $(0,1/2)$. If $y$ isn’t in your sequence, what you need is an open interval around $y$ that misses every multiple of $1/2$.2012-09-28
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    but as you said, it's enough to find an open interval that have at most 1 element of the sequence. we don't need an interval with no elements of the sequence. Am I right?2012-09-28
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    @user42912: True, though it’s more elegant to find one that misses the sequence entirely when the point isn’t in it. However, $\left(y-\frac12,y+\frac12\right)$ still doesn’t quite work, because it contains an integer. That means that it contains the identification point, but it doesn’t contain an open neighborhood of that point. If you’re just going to find a nbhd that contains at most one point of the sequence, use $(\lfloor y\rfloor,\lfloor y\rfloor+1)$ when $y\notin\Bbb Z$.2012-09-28
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    yes, I found a more elegant neighborhood, I have just a question,why a neighborhood with an integer doesn't work? why it doesn’t contain an open neighborhood of that point?2012-09-29
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    @user42912: Because the identification point in $Y$ corresponds to **all** of the integers. A nbhd of that point must contain a nbhd of **every** integer.2012-09-29
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This is Example 1.25: The Shrinking Wedge of Circles on page 49 of Allen Hatcher's book "Algebraic Topology". The fundamental group of the Hawaiian Earrings is uncountable, while the fundamental group of the wedged sum of countably many circles has countably many generators and so is itself countable. Therefore the Hawaiian Earrings and $\mathbb{R}\backslash\mathbb{Z}$ have distinct fundamental groups and thus fail to be homeomorphic.

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    thank you Fly by night, but I suppose to solve this with general topology.2012-09-28