How can a single point "b" be mapped to $0\times [-1,1]$?

Question

I've easily shown that Topologist's Sine Curve(TSC) is connected.Now I'm trying to fill the gaps in the following proof of "showing TSC is not path-connected" .While going through this proof i got some questions,i was unable to answer .

Suppose there is a path $f:[a,c]\rightarrow \bar S,$ from $(0,0)$ to (1,0) in the Topologist's Sine Curve(TSC).

Consider the set of those $t$ for which $f(t)\in 0\times [-1,1]$ is closed,let this set be $K$.So,K is closed(K contains all of its limit points).So, K has a largest element 'b'(say) ({1}).( check the arguments in this paragraph,if there is some scope of improvement please let me know )

Then $f:$[b,$c$]$\rightarrow \bar S$ is a path that maps b into the vertical interval {$0$}$\times [-1,1]$ and maps other points of [b,c] to points of $S$.({2},{3})

We relabel the domain as $[0,1]$ for convenience.

let $f(t)=(x(t),y(t))$, where $y(t)=\sin({\frac{1}{x(t)}})$

As $f$ is continuous,the coordinates $x(t)$ and $y(t)$ are continuous as well.

We produce a sequence $<t_n>\rightarrow 0$ in $[0,1]$ such that $y(t_n)\not\to y(0)$,which violates the sequential criteria for convergence in Metric spaces.

Since,$f:[0,1]\rightarrow \bar S$ is a path.

So,$f(0)=x(0)$ and $f(t)=y(t)$ for $t>0$.

But,$f(0)=0$,So,$x(0)=0$ and $f(t)=y(t)$ for $t>0$.

Now,$x(0)=0$ and $y(t)=\sin (\frac{1}{x(t)})$ for $t>0$.

To find $<t_n>$,we proceed as follows:

Given $n$,choose $u$ with $0<u<x(\frac{1}{n})$ such that $\sin(\frac{1}{u})=(-1)^n$.

Since,$x(t)$ is continuous on $[0,1]$.So, by intermediate value theorem there exists $t_n\in (0,\frac{1}{n})$ such that $x(t_n)=u$ i.e.,$\sin(\frac{1}{u})=\sin(\frac{1}{x(t_n)})=y(t_n)=(-1)^n.$({4})

query from the above proof-

(2) How can a single point "b" be mapped to $0\times [-1,1]$?

Reference:Topology by J.R.Munkres(2nd ed.)

Answer 1

How can a single point "b" be mapped to $0\times [-1,1]$?

This certainly is not what the author meant. What he wanted to say is simply $f(b)\in \{0\}\times [-1,1]$.

Answer 2

Here are some thoughts:

1. Small typo: $\{ 0 \}$ and $[0,1]$ are not subsets of $\Bbb{R}^2$.
2. $K$ does not have a largest element. It is a subset of $\Bbb{R}^2$, which doesn't have a standard ordering on it. I think you're looking for a largest $t$ such that $f(t) \in K$. But such a thing may also not exist, you should instead be thinking about the supremum / least upper bound.
3. To answer your main question: the image of $\bf b$ isn't the whole set $K$; the wording $\bf f$ is mapped into $K$ is important here. It just means that $f({\bf b})\in K$.

I hope this helps!