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### Function Open but Not Continuous

```Date: 03/06/2003 at 20:02:56
From: Karsten
Subject: Find a map R to R that is open but not continuous

Find a map R to R that is open but not continuous.

In trying to find a map that is not continous, there are many
continuous functions that are open.  I think I have found one; let me
know if it is close.

Let X = Y = R. Let Fx denote the discrete topology on X and Fy denote
the usual, metric topology on Y. Define F:(Y,Fy) to (X, Fx) by F(x)
= x. Then g is open, since the image of any set is open in (X, Fx),
but f is not continuous since the inverse image of the open set {0}
is not open in (Y, Fy).
```

```
Date: 03/07/2003 at 07:39:04
From: Doctor Jacques
Subject: Re: Find a map R to R that is open but not continuous

Hi Karsten,

You example is correct; however, I think you are supposed to find such
a function using the normal topology on both sets.

Note that the opposite problem is easy: to find a continuous but not
open function, consider, for example, f(x) = x^2. The image of the
open interval (-1,+1) is not an open set.

Let us now go back to the problem at hand.

Such functions are hard to find, and the example I will show you is
pretty weird.

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).

As I said, this is a pretty weird example.

some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/07/2003 at 14:45:31
From: Karsten
Subject: Thank you (Find a map R to R that is open but not continuous)

Hello Dr. Math,

Thank you for responding to my question. I should have stated that
this is a practice problem given by my instructor and I wanted to
see if I was close to a solution and then compare it to my
instructor's example. Also, the problem seemed really interesting to
me becuase there are many functions that are continuous and open but
many that are not continuous and open, and it is much harder to find
to find such functions.

I'm studying math, and I really enjoy the subject and hope to teach
math in the future. Thanks again for responding.
- Karsten
```
Associated Topics:
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