Subspace topology and homeomorphic mapping The following example is written in Set Theory and General Topology (by Fuichi Uchida, Textbook in the Japanese Language).

Example 25.4 Let subspace $X$ of $R^2$ with usual topology be
$X=\{(x,\sin(\frac{1}{x}))\mid x>0\} \cup \{(0,y) \mid -1\leq y \leq 1\}.$
Let subset A of X be
$A=\{(x,\sin(\frac{1}{x}))\mid x>0 \}.$
We have $A$ is homeomorphic to the interval $(0,\infty)$.

I would like to know a proof that $A$ is homeomorphic to the interval $(0,\infty)$. We need to show that the mapping $(0, \infty) \rightarrow A$ is bijective and continuous, and that the inverse mapping is also continuous. The mapping $(0, \infty) \rightarrow A$ is clearly bijective. If we could accept that $\sin(\frac{1}{x})$ is a continuous function for $x > 0$, then we know that this mapping is continuous. Finally, we should prove that the inverse mapping $A \rightarrow (0, \infty)$ is continuous. However, I could not find the proof. Please give me hints concering the inverse mapping $A \rightarrow (0, \infty)$ is continuous.
PD. Sorry about my English, but it is not my native language.
 A: Consider$$\begin{array}{rccc}f\colon&(0,\infty)&\longrightarrow&A\\&x&\mapsto&\left(x,\sin\left(\frac1x\right)\right).\end{array}$$Then $f$ is a continuous bijection. And $f^{-1}\colon A\longrightarrow(0,\infty)$ is continuous too, since it is simply the map $(x,y)\mapsto x$. Therefore, $f$ is a homeomorphism.
A: $A$ is the graph of the continuous map $\sin(\frac 1 x) : (0,\infty) \to \mathbb R$.
We shall prove the following theorem:

The graph $G(f) = \{(x,f(x)) \mid x \in X \} \subset X \times Y$ of a continuous map $f : X \to Y$ is homeomorphic to the domain $X$ of $f$. Here $G(f)$ receives the subspace topology of the product $X \times Y$ (which is endowed with the product topology).


*

*The map $i : X \to G(f), i(x) = (x,f(x))$, is continuous.


*The projection $p_X: X \times Y \to X$ is continuous, thus also the restriction $p = p_X \mid_{G(f)} : G(f) \to X$ is continuous.


*We have $p \circ i = id_X, i \circ p = id_{G(f)}$. Thus $i$ and $p$ are homeomorphisms which are inverse to each other.
