What does $[0,1]/\{\frac 1n : n \in \mathbb N \cup \infty\}$ look like - where the quotient operation is the quotient topology after identifying the latter set with a point?
I'm mostly curious if this space can be visualized. Any open set in $[0,1]$ that doesn't contain one of $\{\frac 1n : n \in \mathbb N \cup \infty\}$ is unchanged by the quotient operation - but any open set that contains one of these points then contains all of them, and in fact contains every point in the interval close enough to $0$.
My naive guess is that this space is homeomorphic to the Hawaiian earring - each $[\frac 1{n+1}, \frac 1n]$ becomes identified with a circle, and these become "increasingly smaller" as $n$ decreases (this is topologically nonsense, but I'm not sure how to word it in a way that's not, if this is even possible). Am I right in my guess? If so, how would I go about proving the existence of a homeomorphism?
The Hawaiian Earring $H$ is usually defined as the subspace of $\Bbb R^2$ which is the union of circles $C_n$ where $C_n$ has radius $1/n$ and center $(1/n,0)$.
Consider the maps $f_n:\left[\frac1{n+1},\frac1n\right]\to H$ defined by $$f_n(x)=\left(\frac1n,0\right)+\frac{e^{(2n(n+1)\pi i(x-1/(n+1)))}}n$$ Each $f_n$ maps its domain onto the circle $C_n$, identifying only the end points of the interval. Now consider $f:[0,1]\to H$ defined by $$f(x)=\begin{cases} f_n(x), &\text{ if }1/n\ge x\ge 1/(n+1)\\ (0,0), &\text{ if }x=0 \end{cases}$$
This $f$ has a continuous restriction on $(0,1]$ since the domains of the $f_n$ form a locally finite family of closed sets. As $(0,1]$ is open in $[0,1]$, $f$ is continuous on $(0,1]$. I leave the prove of the continuity at $0$ to you. This $f$ now induces a continuous bijection $\tilde f:[0,1]/\sim\to H$, which is a homeomorphism since the domain is compact and $H$ is Hausdorff.