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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"?
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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.)

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    $\begingroup$ 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. $\endgroup$ – StumpyLeg Jun 17 '14 at 18:55
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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.

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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.

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

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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.

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