# how we show that the hawaiian earrings are not homeomorphic to this space

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

-

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

-
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 subsequence – user42912 Sep 28 '12 at 22:15
@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? – Brian M. Scott Sep 28 '12 at 22:19
Let y be not an integer, just take this neighborhood (q-1/2, q+1/2), am I right? – user42912 Sep 28 '12 at 22:40
@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 $1/2$. – Brian M. Scott Sep 28 '12 at 22:44
@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. – Brian M. Scott Sep 29 '12 at 20:43

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.

-
thank you Fly by night, but I suppose to solve this with general topology. – user42912 Sep 28 '12 at 22:12