I was reading Munkres' proof that topologist's sine curve is not path connected. What confused me was why he bothered to reduce the path $f:[a,c]\to\bar{S}$ to $f:[b,c] \to \bar{S}$. As is understood by me, it is to assure that $x(t)\gt0$ for $t>b$, so that one can always choose $u$ with $0 < u < x(1/n)$ for arbitrary large $n$. To put it in another way, the crucial point is to use the right discontinuity of the point $b$. Is that so? What if one proceed with the entire path instead? Thanks in advance!
Best Answer
From a purely logical standpoint, Munkres' reduction is not needed.
Let $g(x)=\sin(1/x)$, $0<x\leq 1$. Let $S(x)=(x,g(x))$, the graph of $g$. The key fact is that for any $x_0>0$, there are infinitely many values of $x$ between $0$ and $x_0$ with $g(x)=1$, and with $g(x)=-1$.
Say $f:[a,c]\to\overline{S}$ is a path connecting $(0,0)$ to some point in $S$. Let $f(t)=(x(t),y(t))$ with $x(a)=0$ and $x(c)=x_0>0$. Set $t_0=c$, and inductively define a sequence $t_0>t_1>\ldots >a$ with $x(t_n)=x_n$ and $g(x_n)=(-1)^n$ for all $n\geq 1$. The mere existence of such a sequence shows that $t\mapsto g(x(t))$ cannot be a continuous function, since $t_n$ must have a limit greater than or equal to $a$, but $g(x(t_n))$ cannot have a limit. If $t\mapsto g(x(t))$ is not continuous, then $t\mapsto (x(t),g(x(t))=f(t)$ isn't either.
The sequence $\{t_n\}$ can be defined inductively. Suppose $t_{n-1}$ has been defined. We know that there is an $x$-coordinate, say $u$, such that $g(u)=(-1)^n$ and $0<u<x_{n-1}$. Now we just need a time-coordinate to go with it: a $t_n$ such that $a<t_n<t_{n-1}$ and $x(t_n)=u$. But $u$ is between $0=x(a)$ and $x_{n-1}=x(t_{n-1})$, so by the intermediate value theorem, $x(t_n)=u$ for some $t_n$ inbetween $a$ and $t_{n-1}$.
This is basically Munkres' argument. The fact that $f$ might not move for awhile--that $f(t)$ might be stationary between $a$ and some $b>a$--turns out to be irrelevant to the logic. However, I do think that Munkres' argument is more appealing from a visual standpoint.