The following is a proposed proof.
Topology's Jordan curve theorem, first proposed in 1887 by Camille Jordan, asserts that any simple closed curve—that is, a continuous closed curve that does not cross itself (now known as a Jordan curve)—divides the plane into exactly two regions, one inside the curve and one outside, such that a path from a point in one region to a point in the other region must pass through the curve.
Proof of Jordan's assertion was considered to be important because of the existence of non-differentiable curves, such as Koch's Snowflake.
The following proof uses an axiom of standard set theories (ZFC or NBG) to assure that all continuous real curves are included. Among those are curves associated with non-computables, which cannot be visualized or even algorithmized.
The key axiom is the well-ordering axiom, which says that every pure set has a least element. Hence the set of reals is well-ordered, including the subset of non-computables.
Now it is straightforward that any finite set can be ordered in n! ways. But what of infinite sets? If we accept the existence of infinite sets -- as nearly all mathematicians do -- then we ought accept that any infinite set can be reordered -- even if we cannot apprehend that permutation. Of course some orderings may be questionable. How would one put the "last" member of an infinite set first if the last number doesn't exist? But clearly a great many other orderings are acceptable: as in 2,1,4,3,6,5...
Or consider the first element ε of R in (0,1]. The well ordering theorem requires that ε exist, but good luck finding it!
So, we shuffle R on (0,1] in all permissible ways and don't worry about whether some shuffles are impermissible.
We then map R2 onto the x-y grid, and use the first and fourth quadrants.
We write functions f(x) = y, where y is the height in Quad I and g(x) the negative height in Quad II.
We apply f to R and map it to its corresponding point in R' , whereby R' is bijective with R but is an arbitrary ordering of R. Similarly for R'' mapped into Quad II.
We require that the plane points denoted 0,0 and x,0 intersect the Quad I and Quad II curves.
At this juncture, we see that the arbitrary curve is continuous in x and possibly wholly or partly discontinuous in y, since yb - ya may be a finite, "non-infinitesimal" distance.
Now a curve like x2 is continuous in x but might seem discontinuous in y though differentiable in y. The curve itself is continuous no matter what our function but the heights, as said, aren't always so.
In any event, whether we say the y heights are discontinuous or continuous, or both, is interesting but irrelevant. What matters is that the x component is continuous and that the y heights exist, at least platonically or (:D) trans-platonically.
Now an immediate consequence of the well-ordering axiom is that any three numbers can be ordered such that a < b < c. So we say that all y ordinates [0, f(x) ) are inside the curve. All numbers greater than f(x) are outside the curve. The same reasoning is applied to Quad II, though we avoid a trivial redundancy by writing (0, g(x) ).
One may wonder whether a continuous path might cut through the f(x) curve where two adjacent y heights are discontinuous. But the path is itself a curve defined by both x and y components. That is, even if there were room for a path to squeeze through vertically, there is no squeeze room horizontally.
Still, a curve on some plane that intersects the R2 plane only in the y's, could penetrate f(x) or g(x), but of course only partially.
QED
Topology's Jordan curve theorem, first proposed in 1887 by Camille Jordan, asserts that any simple closed curve—that is, a continuous closed curve that does not cross itself (now known as a Jordan curve)—divides the plane into exactly two regions, one inside the curve and one outside, such that a path from a point in one region to a point in the other region must pass through the curve.
Proof of Jordan's assertion was considered to be important because of the existence of non-differentiable curves, such as Koch's Snowflake.
The following proof uses an axiom of standard set theories (ZFC or NBG) to assure that all continuous real curves are included. Among those are curves associated with non-computables, which cannot be visualized or even algorithmized.
The key axiom is the well-ordering axiom, which says that every pure set has a least element. Hence the set of reals is well-ordered, including the subset of non-computables.
Now it is straightforward that any finite set can be ordered in n! ways. But what of infinite sets? If we accept the existence of infinite sets -- as nearly all mathematicians do -- then we ought accept that any infinite set can be reordered -- even if we cannot apprehend that permutation. Of course some orderings may be questionable. How would one put the "last" member of an infinite set first if the last number doesn't exist? But clearly a great many other orderings are acceptable: as in 2,1,4,3,6,5...
Or consider the first element ε of R in (0,1]. The well ordering theorem requires that ε exist, but good luck finding it!
So, we shuffle R on (0,1] in all permissible ways and don't worry about whether some shuffles are impermissible.
We then map R2 onto the x-y grid, and use the first and fourth quadrants.
We write functions f(x) = y, where y is the height in Quad I and g(x) the negative height in Quad II.
We apply f to R and map it to its corresponding point in R' , whereby R' is bijective with R but is an arbitrary ordering of R. Similarly for R'' mapped into Quad II.
We require that the plane points denoted 0,0 and x,0 intersect the Quad I and Quad II curves.
At this juncture, we see that the arbitrary curve is continuous in x and possibly wholly or partly discontinuous in y, since yb - ya may be a finite, "non-infinitesimal" distance.
Now a curve like x2 is continuous in x but might seem discontinuous in y though differentiable in y. The curve itself is continuous no matter what our function but the heights, as said, aren't always so.
In any event, whether we say the y heights are discontinuous or continuous, or both, is interesting but irrelevant. What matters is that the x component is continuous and that the y heights exist, at least platonically or (:D) trans-platonically.
Now an immediate consequence of the well-ordering axiom is that any three numbers can be ordered such that a < b < c. So we say that all y ordinates [0, f(x) ) are inside the curve. All numbers greater than f(x) are outside the curve. The same reasoning is applied to Quad II, though we avoid a trivial redundancy by writing (0, g(x) ).
One may wonder whether a continuous path might cut through the f(x) curve where two adjacent y heights are discontinuous. But the path is itself a curve defined by both x and y components. That is, even if there were room for a path to squeeze through vertically, there is no squeeze room horizontally.
Still, a curve on some plane that intersects the R2 plane only in the y's, could penetrate f(x) or g(x), but of course only partially.
QED
