# Is $0$ the midpoint of $(-\infty,+\infty)$?

Is $0$ the midpoint of $(-\infty,+\infty)$?

Intuitively, I'd think so, and trying to refine my intuition as to why I'd think so, I would say that this is the case because there is a one-to-one correspondence between $(-\infty,0)$ and $(0,+\infty)$. So there's an equal amount of numbers on either side.

On the other hand, there's a one-to-one correspondence between $(-\infty,k)$ and $(k,+\infty)$ for all $k\in\mathbb R$, which suggests that any real number is a good midpoint. This obviously does not match the intuition for what I mean by "midpoint".

So, the two questions are:

1. Is $0$ a midpoint of $(-\infty,+\infty)$?
2. If the notion of midpoint is inconclusive or vague when applied to $(-\infty,+\infty)$, is there another notion I could apply which captures my intuition about $0$ being the "midpoint"?

For bounded intervals, the midpoint is the centre of symmetry: reflection in $p$ is the function $x\mapsto 2p-x$, and the interval $[a,b]$ is symmetric under this operation with $p=\frac{a+b}2$. The interval $(-\infty,\infty)$, however, is symmetric under this operation for any $p$, so all points are midpoints in this sense.

One high-falutin way to say this is that "midpoint" in the sense of "centre of reflective symmetry" is a notion of affine geometry, and the real line qua affine space has no preferred origin. In affine geometric terms, there's nothing special about $0$. To recognize the specialness of $0$, we need to take into account more of the structure of the real numbers. (For example, the real numbers qua additive group are symmetrical under reflection in $0$, but not under reflection in other points.)

• One could capture the OP's intuition by saying $0$ is the symmetry point for the absolute-value function. This is basically Steven's point, I take it: one has to have a context that makes $0$ special, within which one can then identify something uniquely "middle-like" about it. – StumpyLeg Jun 17 '14 at 18:55

In a good sense a midpoint needs to be definable somehow. But this is a problem if you only consider $\Bbb R$ as an ordered set.

The reason is that, as you said, there are automorphisms (order preserving bijections) which move $0$ to any other element. So if $0$ would satisfy the definition of a midpoint, any other element would have to as well.

So either all points are midpoints, or no points are midpoints. This is up to you, but it seems to me that a midpoint should be unique, to some extent anyway, which doesn't hold here.

I'd argue that there is no well-defined midpoint. The definition I use for midpoint is with regards to any interval $(a,b)$ (along with the closed intervals or half-open intervals): $c$ is the midpoint of $(a,b)$ if $c-a=b-c$.

Now, if we try and do the same with regards to $(-\infty,\infty)$, we must use subtraction and addition of $\pm \infty$. Allowing ourselves to do so (with the common definitions $\infty+r=\infty$ and $r-\infty=-\infty$ for $r$ real, as is typically done with regards to the extended real line) we have that $c-(-\infty)=c+\infty=\infty$ and $\infty-c=\infty$, with $c$ and arbitrary real number. Thus, we find that there can be no unique midpoint of $(-\infty,\infty)$ in the traditional sense.

Another approach would be to use the fact that the midpoint is the centroid of a line segment of uniform density/mass, in which case $c=\int_a^b{xdx}=\frac{b-a}{2}$. But attempting to do the same for $(-\infty,\infty)$ gives rise to the integral $\int_{-\infty}^\infty{xdx}$ which does not converge. However, in spite of this, if you find the Cauchy Principal Value of the integral, you find 0.

Everywhere is the "middle" if you can indefinitely go in either (any) direction.

0 is the midpoint of -N and N for all real numbers N. I don't know what "midpoint" would mean when we have no reference points.