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From Wikipedia, "Geometric points do not have any length, area, volume or any other dimensional attribute. A common interpretation is that the concept of a point is meant to capture the notion of a unique location in Euclidean space."

Suppose it shrinks by $1/2$ scale factor, so $\{{1/2^{n}}\}_{n=0}^\infty$ converges to point $0$. Does it have its Euclidean "square information" replaced with "location information"? At that point (pun intended), is unshrinking it back to its original shape irreversible? Thanks.

half-shrink

This idea came up when I was looking at the inscribed square problem and thought about shrinking a square to a point, so every point on the curve contains four vertices of information. That's "cheating". However, as noted in the answer below: "shrinking your square into a single point will make it lose its square information".

squares

vengy
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2 Answers2

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Consider your square to be the set of points $X\subset\mathbb{R}^2$

Consider the function $\varphi:X\to\mathbb{R}^2$ that turns each point of your square into the same single point $(x,y)\in\mathbb{R}^2$. That means, for any point $(a,b)$ from your square, then $$\varphi((a,b))=(x,y).$$ So it's clear that this $\varphi$ isn't injective (multiple points give the same image, all points in fact). Also, it isn't surjective, since only $(x,y)$ has a preimage in $\varphi$. Since $\varphi$ is not a bijection, you can't find any inverse function $\varphi^{-1}$ to undo what you did to your square using $\varphi$.

So yes, shrinking your square into a single point will make it lose its "square information".

Alejandro Bergasa Alonso
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    Good to know...the inverse $\varphi^{-1}$ bijection does not exist, so that explains it. Thanks! – vengy Jan 08 '21 at 15:42
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A rectangle with sides only under consideration, can not be contracted to a point.
There is no point $x_0$ on the sides of the rectangle such that for all x on the rectangle, the line segment from $x_0$ to x completely lies on the rectangle.

From Contractible space and star domain

amitava
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  • Suppose it shrinks by $1/2$ each time as you suggested. The sequence ${{1/2^{n}}}_{n=0}^\infty$ converges to $0$. Once it's $0$, from another answer above, it loses its square property. – vengy Jan 08 '21 at 15:58
  • @vengy , in the link https://en.wikipedia.org/wiki/Contractible_space , which picture at the right you think your picture is similar to ? – amitava Jan 08 '21 at 17:05
  • Wow...so my question doesn't make any sense because it's a non-contactable space to begin with. But, on that same link: "Any Euclidean space is contractible, as is any star domain on a Euclidean space." – vengy Jan 08 '21 at 18:00
  • @vengy If you see the link for the star domain, it says "if there exists an x0 in S such that for all x in S the line segment from x0 to x is in S.", but for a rectangle, when we consider only the sides but not the inside area, it becomes non Euclidean, just like the annulus shown there. – amitava Jan 08 '21 at 19:32
  • @vengy I edited my answer to keep the Wikipedia definition only. It is much broader. – amitava Jan 09 '21 at 04:12