# Why does $b(t)=\text{const}.$ follow from $<b(t),\frac{w}{|w|}\ge \text{const}.$

I want to understand the following proof. It is from the book "Differential geometry of curves and surfaces" by C. Tapp.

I don't understand why $b(t)=const.$ follows from $<b(t),\frac{w}{|w|}>=const.$

That $b(t)$ has to be constant since it continuous and $b(t)=\pm \frac{w}{|w|}$ sort of makes sense, but in the text there is an apparently "more rigorous” way of showing this. The argument uses the fact that a continuous function which attains only integer values is constant. Therefore $<b(t),\frac{w}{|w|}\ge \text{const}.$ , since it is equal to either 1 or -1. From that it is concluded that $b(t)$ is constant, which I don’t understand. Does this follow from $<b'(t),\frac{w}{|w|}\ge 0$? I would like to get something like $b'(t)=0$, because then $b(t)=\text{const}$.

Thank you very much in advance!

Keep in mind that $\mathfrak{b}(t)$ and $\mathbf w$ are parallel and that $\mathbf w$ is a fixed vector. Furthermore, $\bigl\|\mathfrak{b}(t)\bigr\|=1$. Conclusion: $\mathfrak{b}(t)=\pm\frac{\mathbf w}{\|\mathbf w\|}$. Therefore, for each $t$, $\left\langle\mathfrak{b}(t),\frac{\mathbf w}{\|\mathbf w\|}\right\rangle=\pm1$. But the function $t\mapsto\left\langle\mathfrak{b}(t),\frac{\mathbf w}{\|\mathbf w\|}\right\rangle$ is continuous. Therefore, it is either always equal to $1$ or always equal to $-1$. In the first case, $\mathfrak{b}(t)=\frac{\mathbf w}{\|\mathbf w\|}$ for each $t$, whereas in the second case $\mathfrak{b}(t)=-\frac{\mathbf w}{\|\mathbf w\|}$ for each $t$.