Quotient space $[0,1]/C$ is homeomorphic to $[0,1]/\left\{0,1,\frac12,\frac13,...\right\}$, $C$ denotes Cantor set.

Question

How to prove quotient space $[0,1]/C$, where $C$ denotes Cantor set, is homeomorphic to $[0,1]/\left\{0,1,\frac12,\frac13,...\right\}$?

Answer 1

Both complements $[0,1] \setminus C$ and $[0,1] \setminus \left\{0,1,\frac12,\frac13,...\right\}$ are the union of countably infinitely many disjoint open intervals.

Let $H = \bigcup_{n=1}^\infty S(\frac{1}{n};\frac{1}{n})$ be the Hawaiian earring. Here $S(a;r) \subset \mathbb{R}^2$ denotes the circle with center $(a,0)$ and radius $r$. This is a compact subset of the plane. We shall prove

If $A \subset [0,1]$ is a closed subset such that $0,1 \in A$ and $[0,1] \setminus A$ is the union of countably infinitely many disjoint open intervals $U_n =(a_n,b_n)$, then $[0,1]/A \approx H$.

Define $A_n = A \cup U_n$. These are closed subsets of $[0,1]$ because the points $a_n,b_n$ must belong to $A$. There are continuous maps $f_n : A_n \to S(\frac{1}{n};\frac{1}{n})$ such that $f_n(x) = (0,0)$ for $x \in A$ and mapping $(a_n,b_n)$ homeomorphically onto $S(\frac{1}{n};\frac{1}{n}) \setminus \{ (0,0) \}$.

Define $f : [0,1] \to H, f(x) = f_n(x)$ for $x \in A_n$.

$f$ is obviously continuous in all points of $[0,1] \setminus A$. Let us check that is also continuous in all $x \in A$. We have $f(x) = (0,0)$. Let $V$ be an open neighborhood of $(0,0)$ in $H$. Then $S(\frac{1}{n};\frac{1}{n}) \subset V$ for $n \ge N$. Hence $f(A_n) = f_n(A_n) \subset V$ for $n \ge N$. Let us consider $n < N$. The sets $f_n^{-1}(V)$ are open in $A_n$ and can be written as $f_n^{-1}(V) = W_n \cap A_n$ with open neighboorhoods $W_n$ of $x$ in $[0,1]$ (recall $x \in A \subset A_n$). Let $W = \bigcap_{n=1}^{N-1} W_n$. Then $f(W) = f(\bigcup_{n=1}^\infty (W \cap A_n)) = \bigcup_{n=1}^\infty f(W \cap A_n) = \bigcup_{n=1}^\infty f_n(W \cap A_n) \subset V$.

Since $[0,1]$ is compact Hausdorff, $f$ is a closed map and therefore a quotient map. Obviously $f$ induces a continuous bijection $f' : [0,1]/A \to H$ such that $f' \circ p = f$, where $p : [0,1] \to [0,1]/A$ denotes the quotient map. By the universal property of quotient maps $f'$ is a homeomorphism.

Remark:

The $U_n$ are the connected components of $A^* = [0,1] \setminus A$. For an arbitrary closed $A \subset [0,1]$ with $0,1 \in A$ the space $A^*$ cannot have uncountably many components (otherwise the Lebesgue measure of $A^*$ would not be finite). Hence there are either $m < \infty$ components (in which case $[0,1]/A$ is the wedge of $m$ circles which is a point for $m = 0$) or countably infinitely many components (in which case $[0,1]/A$ is a Hawaiian earring).

The number of components of $A^*$ is determined by the number of components of $A$ (these are closed intervals or points). If $A$ has $k < \infty$ components, then $A^*$ has $k - 1$ components. If $A$ has infinitely many components (it may have even uncountably many), then $A^*$ has countably infinitely many components.

If $A$ is nonempty and contains only $r <2$ of the points $0,1 $, then $A^*$ has $2-r$ components which are half-open intervals. In the quotient $[0,1]/A$ they produce $r-2$ copies of $[0,1]$ which intersect $[0,1]/A$ in the equivalence class of $A$.