0
$\begingroup$

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

5

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

  • 1
    @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
2

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.

  • 0
    thank you Fly by night, but I suppose to solve this with general topology.2012-09-28