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This construction I found it paper published in $1965$ I think. Here is the way that defined. Let $I=[0,1]$ and define a Cantor set as follows.

$C_1$ obtained from $I$ by taking the open interval at the center $frac12$ with length $frac14$. In general $C_n$ is the union of $2^n$ closed intervals and $C_n+1$ is obtained from $C_n$ by taking out of each intervals of these $2^n$ intervals a concentric open interval of length $frac12^2n+2$. So, it is easy to see the Lebesgue measure for $C=bigcap_n=1^infty C_n$ is $frac12.$ Now define $f:Irightarrow I^2$ as follows $$ f(t)=(x(t),y(t))$$ where $x(t)=2m(Ccap [0,t])$ and $y(t)=2x(t)-t$ where $m$ is the Lebesgue measure. Let $M=f[I]$ and $K=f[C]$.

Here is my question why $K$ is obtained from $M$ by taking out all open vertical and these occur when abscissa is $frac12$,$frac14$,$frac34$,$frac18$,...

I did not see why this is case. Any help will be appreciated.

Note that $x$ is continuous, increasing (but not strictly), and rises from $0$ to $1$. $y(t)$ is also continuous, but is strictly decreasing wherever $x$ is constant. As a result $f$ is one-to-one. Because $f$ is injective, $$M setminus K = f(I setminus C)$$
Now $I setminus C$ consists exactly of all the intervals removed in the construction of $C$. On each of these intervals, $x$ is constant, and $y$ decreases linearly. Thus $f(t)$ is a vertical line segment on these intervals.

When each of these intervals was removed from $C$, at that step, the two intervals left behind were still unbroken, after which they were treated exactly like $I$ itself. In particular, exactly half of their length is removed. Hence when $t$ is the lower endpoint of a removed interval, $x(t) = 2(frac t2) = t$. So determining the abcissa of these vertical line segments is just a matter of finding the left endpoints of all removed intervals in the construction of $C$.

Finally, one has to show that there are no other vertical line segments, other than when $t$ is in one of the removed intervals. Otherwise, removing vertical segments would remove points from $K$ as well. But the only way to have a vertical segment is for $y$ to change while $x$ is constant, which requires $y$ to change linearly, and therefore over some interval. So $x$ is constant over an interval, which means that the intersection of $C$ and this interval has measure $0$. But from the construction, the only intervals over which $C$ has measure $0$ are subsets of the removed intervals. Therefore no vertical open line segment of $f$ intersects $K$.

Thus $f(I setminus C)$ consists entirely of vertical open lines segments, and $K$ contains no vertical open line segments. So removing the vertical line segments ($f(Isetminus C) = M setminus K$) from $M$ leaves $K$.