I'm looking for $f:\mathbb{R}^2\rightarrow Y$, where $f$ is a continuous surjection in the context of the following problem:
-Show the subspace $Y = \{(x, y) : 0 < x \leq 1, y = \sin (1/x)\}$ of $\mathbb{R}^2$ is connected.
My solution strategy:
I want to apply the theorem that says: Let $f:(X, \tau)\rightarrow (Y,\tau_1)$ be a surjective, continuous function from one topological space to another. Then, if $X$ is connected, so is $Y$.
Question: Will the following function work:
$$f(x,y)= \left\{ \begin{array}{lcc} (x, \sin(1/x)), & 0 < x \leq 1\\ \\ \emptyset , &\mbox{ otherwise,i.e., not defined} \\ \end{array}\right.$$
Clearly, this is a surjection since each element in $Y$ has an inverse image in $\mathbb{R}^2$.
But I'm unsure about how to rigorously establish that $f$ is continuous by means of the general topological definition of a continuous function as a map from open sets to open sets.
Intuitively, I can envisage a correspondence between open rectangles in $\mathbb{R}^2$ whose intersection with the infinite vertical strip $(0,1]\times(-\infty, \infty)$ is mapped vertically to a connected portion of the curve $y=\sin(1/x)$, and the inverse map where connected portions of the curve are mapped to the infinite vertical strip that covers its x-values. But this seems vague and hand-wavy.
Am I on the right track?
[P.S.: Apologies for one more question relating to the Topologist's Sine Curve, but I believe this one is different in its focus on continuity rather than connectedness:
Citation: S. Morris "Topology without Tears", 5.2.6 (i)]
I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.