Now you can see why we wanted to drop \(t=0\) and \(t=1\) from our set corresponding to breathed infinity: the function \(f\) above is undefined at \(t=0\) and \(t=1\).

Next we must convince ourselves that this is a one-to-one correspondence. It is defined on the interval \((0,1)\); in other words, it sends every element of \((0,1)\) somewhere. More specifically, it sends every element of \((0,1)\) to some real number.

**The last thing we need to check is that every element of \(\mathbb{R}\) gets hit by some element of \((0,1)\), using the function \(f\).** The graph of \(f\) goes up higher and higher as the input moves closer to \(1\). The tangent function \(\tan(x)\) has a vertical asymptote at \(\frac{\pi}{2}\) (because tangent is \(\sin x\, / \cos x\), and \(\cos x\) approaches zero as \(x\) approaches \(\frac{\pi}{2}\).) The function \(f\) is a shift and compression of \(\tan(x)\), so it has a vertical asymptote at \(t=1\).

This means that \(f(x)\) does in fact get arbitrarily high, as \(x\) approaches \(1\). For any positive real number \(z\), no matter how large, on the y-axis, if we draw a horizontal line there, the graph will reach up to cross it. In other words, there is some number \(w\) in \((0,1)\) so that \(f(w) = z\).

The same is true for negative real numbers: the function \(f\) goes arbitrarily far down as its input approaches \(0\). So we conclude that for every number \(z\) in \(\mathbb{R}\), there is one and exactly one \(w\) in \((0,1)\) such that \(f(w) = z\). Therefore \(f\) is a one-to-one correspondence between \((0,1)\) and \(\mathbb{R}\).