Can you find a map R to R that is open but not continuous?
以下の文章がところどころ意味がわからなくて、完全に理解したくてもなかなかできないでいます。この英文の主張を解説していただけないでしょうか？よろしくお願いします。
[原文]
As R is homeomorphic to the open interval (0,1), we will define a function f from R to (0,1) that satisfies the requirements. You can then extend it to a function from R to R by composing it with, for example, h(x) = (2x-1)/(x(1-x)) or a similar function.
We compute x in the following way:
* First, drop the integer part of x - our function will be periodic.
* Next, write x in decimal representation. If you have the choice of two representations, for example, 0.5 = 0.4999..., choose the first one.
* Find the last digit '9' in the number (if there is no last 9, see below).
* Remove all the digits up to and including that '9', and consider the remaining digits. As there will be no '9' among them, you can interpret them as the "decimal" expansion of a number in base 9 - that is your f(x), and it will be between 0 and 1.
However, a few things may go wrong:
* There may be no 9 at all in the number, or an infinite number of 9 (not necessarily contiguous).
* The resulting base 9 number may be 0 or 0.888... = 1 (in base 9)
In both of these cases, set f(x) = 0.5 (or any number between 0 and 1).
For example, to get f(3.1952945):
* Drop everything up to the last 9.
* The remaining number is 0.45, which, interpreted as a number in base 9, is 41/81.
We claim that the image of any open interval is the whole open
interval (0,1), which is an open set. This will show that f is open.
Indeed, given an interval I = (a,b), we can find a sub-interval J of length 10^(-k) contained in it, for a suitable k > 1, such that the interval fixes the first (k-1) digits after the decimal point.
For example, the interval I = (1.14567, 1.15321) contains the sub-interval J = (1.151, 1.152).
Given any y in (0,1), we can find an x in J (and therefore in I) such that f(x) = y. To do that:
* write y in base 9
* append a digit 9 to the lower bound of J.
* append the base 9 representation of y.
For example, with I and J as above, we can find an x such that
f(x) = 1/3.
* 1/3 in base 9 is 0.3.
* The lower bound of J is 1.151.
* the number x = 1.15193 satisifies f(x) = y, and belongs to J.
Note that there are many other possible numbers, obtained by choosing smaller intervals for J, for example, 1.1512393, and so on.
It should also be obvious that f is not continuous (in fact it is in some sense the most discontinuous possible function).
Indeed, a function f is continuous at a if, given any open interval K = (f(a)-e,f(a)+e), we can find an open interval J = (a-d,a+d) such that the image of J is a subset of K. But, by what we just saw, the image of any interval is the whole of (0,1).
お礼
お礼が遅くなりすみません。vertical line test やってみました。 ｘ軸に平行な線を引いた時完璧に２点で交わります。これではD f(x)={ -2 ≤ x ≤ 1.14,,,, } は絶対に駄目ですね。 頭の中で考えている事と実際に書いて目で見るのとでは全く違いますね。 有難うございました！