<s>[INST] Is $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$ closed? Injective? Surjective? Let $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$.


*

*Is $x*y$ closed in $(0, \infty)$?

*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty) \to (0,\infty)$ injective?

*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty) \to (0,\infty)$ surjective?

 [/INST] What I tried:


*

*Yes.


If $x,y > 0$, then $x*y > 0$.
QED


*Yes


Let $y_1,y_2 > 0$ s.t. $f(y_1) = f(y_2)$.
Then
$$\frac{1}{\frac{1}{x} + \frac{1}{y_1}} = \frac{1}{\frac{1}{x} + \frac{1}{y_2}}$$
$$\to \frac{1}{x} + \frac{1}{y_1} = \frac{1}{x} + \frac{1}{y_2}$$
$$\to \frac{1}{y_1} = \frac{1}{y_2}$$
$$\to y_1 = y_2$$
QED


*No


Let $a \in \text{codomain}(f) = (0,\infty)$. Suppose $\exists y \in \text{domain}(f) = (0, \infty)$ s.t.
$$\frac{1}{\frac{1}{x} + \frac{1}{y}} = a$$
$$\to \frac{1}{x} + \frac{1}{y} = \frac{1}{a}$$
$$\to \frac{1}{y} = \frac{1}{a} - \frac{1}{x}$$
$$\to \frac{1}{y} = \frac{x}{xa} - \frac{a}{xa}$$
$$\to \frac{1}{y} = \frac{x-a}{xa}$$
$$\to y = \frac{xa}{x-a}$$
Thus, there is no such $y$ if $a=x$.
QED
</s>