finding an injective function between real numbers to prove equal cardinality

Question

A problem in discrete math states that we need to prove that $(0,2010)$ has the same cardinality as $(0,\infty)$. The suggested solution is to find an injective function from $(0,2010)$ to $(0,\infty)$ and vice versa. I thought of the following function: $$f(x)=\frac{1}{x}$$ however the suggested function in the solution is: $$f(x)=\frac{1}{x}-\frac{1}{2010}$$ Why isn't my function enough why do we have to substract the $\frac{1}{2010}$ part?

Answer 1

Your function establishes a bijection between $(0,1)$ and $(\color{red}1, \infty)$ - not what you are asked to show. But if you subtract that annoying $\color{red}1$, i.e. if you consider $\dfrac 1 x - 1$, then you get your desired bijection.

I don't get the $2010$ part and I wouldn't bother with it.

Answer 2

You function shows that $]0,1[$ has the same cardinality as $]1,+\infty[$, $f(x)=\frac{1}{x}-1$ seems like a good answer, and $f(x)=\frac{1}{x}-2010$ would show that $]0,1[$ is bigger than $]0,+\infty[$ (it spans a bigger interval) which is fine since you also have the inclusion the other way.

I don't know why they retracted $\frac{1}{2010}$, it doesn't show anything (it shows it for $]1-\frac{1}{2010},+\infty[$, but that is not the topic discussed here)

Answer 3

Your function bijectively maps $(0, 1)$ to $(1, \infty)$, not to $(0, \infty)$. So, I would put $$ f(x) = {1 \over x} - 1. $$ No idea about the $2010$ denominator.