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 $Y$.
 A: 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.
A: 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.
