Show that a subset $X_j$ of the unit circle with $j$ distinct points removed, has exactly $j$ path components ($j \geq 1$)

Question

Show that a subset $X_j$ of the unit circle with $j$ distinct points removed, has exactly $j$ path components ($j \geq 1$).

I can visually and intuitively see why this is true, but am having trouble providing a more exact explanation.

For $j=1$ it is obvious that $X_j$ will still be path connected.

For $j = 2$, say the removed points are $r_1$ and $r_2$. Then we will have two paths, say $\gamma_1 : [0,1] \longrightarrow X_j$ which connects points in $X_j$ from $r_1$ to $r_2$ clockwise around $X_j$ and $\gamma_2 : [0,1] \longrightarrow X_j$ which connects points in $X_j$ from $r_2$ to $r_1$ clockwise. (Here by "from" I mean points in $X_j$ arbitrarily close to, but not equal to $r_1$ or $r_2$). These functions must be paths since their respective images are subsets of the image of $\gamma_0$ where $\gamma_0$ is the path connecting points in $X_1$. Hence $X_2$ has 2 path components.

Now for some $k > 2$, the removed point $r_k$ from $X_k$ must necessarily be between some $r_x$ and $r_y$ removed from $X_{k-1}$ since the removed points are distinct by assumption. The path connecting points in $X_{k-1}$ from $r_x$ to $r_y$ will now be discontinuous at $r_k$ and is no longer a path in $X_k$. The removal of $r_k$ now creates two paths, say $\gamma_x$ from $r_k$ to $r_x$, and $\gamma_y$ from $r_k$ to $r_y$. Hence $X_k$ will have 1 more path component than $X_{k-1}$, so therefore if $X_{k-1}$ has $k-1$ path components, $X_k$ has $k$ path components.

The result that $X_j$ has exactly $j$ path components for $j \geq 1$ follows by induction. $\Box$

Does this proof contain any errors? How can it be made more precise?

Answer 1

Your idea is nice, but it is not properly elaborated. In fact you say

Then we will have two paths, say $\gamma_1 : [0,1] \longrightarrow X_j$ which connects points in $X_j$ from $r_1$ to $r_2$ clockwise around $X_j$ and $\gamma_2 : [0,1] \longrightarrow X_j$ which connects points in $X_j$ from $r_2$ to $r_1$ clockwise. (Here by "from" I mean points in $X_j$ arbitrarily close to, but not equal to $r_1$ or $r_2$).

One can understand what you mean, but it is rather vague.

I suggest to do it as follows:

1. $X_1$ is clearly homeomorphic to the open interval $(0,1)$. Let $h : X_1 \to (0,1)$ be a homeomorphism.

2. For $j > 1$ we have $X_j = S^1 \setminus \{r_1,\ldots,r_j\} = X_1 \setminus \{r_2,\ldots,r_j\}$, thus $X_j \approx h(X_1 \setminus \{r_2,\ldots,r_j\}) = (0,1) \setminus \{ h(r_2),\ldots, h(r_n)\}$. But now you can easily show by induction that $(0,1) \setminus \{ h(r_2),\ldots, h(r_n)\}$ is the the union of $j$ disjoint open subintervals $I_k \subset (0,1)$. Each $h^{-1}(I_k)$ is a path component of $X_j$.