Infinite Earring and Product Space Homeomorphic

Old qual question here:

We define two topological spaces $X$ and $Y$ as subspaces of certain topological spaces. $X$ is defined as a subspace of $\mathbb{R}^2$ which is the union of the infinite number of circles: $$X=\bigcup_{i=1}^\infty S^1_n\subset\mathbb{R}^2$$ where $S_n^1$ is a circle in $\mathbb{R}^2$ with radius $1/n$ and center at $(1/n,0)$. Now let $S^1$ denote a unit circle with a marked point $p_*$. $Y$ is defined as a subspace of the infinite product of circles $\prod_{i=1}^n S^1$ with product topology, consisting of points $(p_1,p_2,\dots)$ where $p_i=p_*$ for all $i$ except one. Is $X$ homeomorphic to $Y$?

We are very stuck. We only have point set topology at this point in the course. Our thoughts include $\overline{X}=X$ wheras $\overline{Y}$ contains the point $(p_*,p_*,\dots)\not\in Y$, but we are not sure how to capture that idea with homeomorphism (since closure depends on an ambient space). We also know that $X$ is compact, not sure about $Y$.

• $Y$ is not compact! Indeed any compact subset of a Hausdorff space must be closed, but you've shown it is not equal to its own closure in the countable product of circles. Feb 29, 2016 at 21:29
• Are you certain it explicitly says $p_i = p_*$ for all $i$ except one? Are you sure it didn't say for all $i$ except for at most one? Feb 29, 2016 at 21:32
• @GrumpyParsnip great answer, thank you. Mar 1, 2016 at 15:22
• @MartianInvader yes that is exactly what it says Mar 1, 2016 at 15:22

If $Y$ does not include the point $(p_\ast, p_\ast, \ldots)$, then the answer is certainly no: $X$ is compact while $Y$ is not.

If $Y$ is supposed to include this point, however, then the answer is yes. To see this, note that both $X$ and $Y$ are a one-point compactification of the disjoint union of countably many open intervals.

The result follows from the fact that one-point compactifications of locally compact Hausdorff spaces are unique up to homeomorphism.

• Are you sure that the union of circles inside the countable product is compact, even including the basepoint? Could you not take a smallish neighborhood of the midpoint of each circle together with a neighborhod of the basepoint to get an open cover with no finite subcover? Feb 29, 2016 at 21:46
• From what I can see, the subspace of the infinite product is a wedge of countably many circles, which is well-known to not be compact. I may be missing something though. Feb 29, 2016 at 21:48
• @GrumpyParsnip It's not the wedge of circles, because the product topology is formed via a basis of the product of open sets in each factor, all but finitely many of which are the entire space. So any neighborhood of the basepoint must contain all but finitely many of the circles. The tradional "wedge of circles" generated as a cell complex does not use the product topology. Feb 29, 2016 at 21:51
• @GrumpyParsnip I agree with MartianInvader. To see directly that this is a compact space, note that it's closed in the product topology, and $\prod_{i=1}^\infty S^1$ is a product of compact spaces. Feb 29, 2016 at 21:55
• Ah, of course. Too late in the day for me. Feb 29, 2016 at 22:06

If the question is exactly as you state it, ie, $Y$ is defined as the points $\left(p_1, p_2,...\right)$ where $p_i \neq p_*$ for exactly one i, then $X$ and $Y$ are not homeomorphic for lots of reasons. The easiest is probably that $Y$ is not connected while $X$ is.

If, as I suspect, the point $\left(p_*, p_*, p_*,...\right)$ is included in $Y$, then the two spaces are homeomorphic. To prove this, try proving the following two statements:

1) $X$ can be described as the set of a countably infinite number of circles glued together at a point. The open sets containing that point consist of all but finitely many of the circles (since $X$ inherits its topology from the plane), unioned with open subsets from the remaining (finitely many) circles. Open subsets not containing that point are unions of open subsets of the intervals attached to it.

2) $Y$ can also be described as the set of a countably infinite number of circles glued together at a point. The open sets containing that point must consist of all but finitely many of the circles (since basis sets in the product topology require open sets to contain the full space in all but finitely many factors), and open subsets from the remaining (finitely many) circles. Open subsets not containing that point are unions of open subsets of the intervals attached to it.

Notice that these statements show the topologies on $X$ and $Y$ can be described as the same sets on the same space, thus $X$ and $Y$ are homeomorphic.

• I double checked and the problem is exactly as I wrote it. Could you outline an argument for why it is not connected? I think the hardest part is trying to visualize/understand this space. I don't have a good intuition for it. Mar 1, 2016 at 15:20
• If it helps you visualize, $Y$ is homeomorphic to $X$ with the origin removed. To show it's not connected, let $A$ be the set of points in $Y$ whose first coordinate is not $p_*$, and let $B$ be everything else. Then show that $A$ and $B$ are disjoint open sets that cover $Y$. Mar 1, 2016 at 17:41