SVM: Why do we demand the hyperplanes to be equal $\pm 1$? I'm following this well-known tutorial on SVM.
What I'm trying to understand is why we demand the hyperplanes to be $wx - b = \pm 1$? and why is it always possible and done without loss of generality (I've understood that you can always scale $w,b$ accordingly - is that correct?)
 A: First of all, we demand a hyperplane $H$ to be of the form $\vec{w}\vec{x}+b = 0$ (meaning every $\vec{x}$ that satisfies this condition is part of $H$) such that $\Vert \vec{w} \Vert$ is minimal, under constraints that $\vec{w}\vec{x}+b \ge 1 $ if $f(\vec{x})=1$ and $\vec{w}\vec{x}+b \le -1 $ if $f(\vec{x})=-1$. With $f$ the function that gives the class of a certain example $x$, in this case $1$ and $-1$ (positive and negative example). The last two conditions make sure that examples are at the correct side of the hyperplane and this is where $\pm 1$ originates.
The above minimisation problem results in a value for $w$ and $b$ in the 1-D case that maximises the margin between positive and negative examples and placed the hyperplane at the optimal position. The sideview in the slide below might clear things up as you should not confuse the hyperplane with the red line, the latter just represents the values of $wx+b$ for each $x$. The actual hyperplane consists of every $x$ that satisfies $wx+b=0$, in the 1-D case just a single point.
And we thus do not demand the hyperplanes to be $wx+b=\pm 1$

