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 \(\mathbb{R}\).