1
$\begingroup$

Let $S$ be a set of continuous functions defined on an interval $[0,1]$. The addition and multiplication between two elements is defined as $(f+g)(x)=f(x)+g(x)$ and $(f\cdot g)(x)=f(x)g(x)$. So $S$ is a commutative ring with unity. Show only idempotents are $0,1$; which are constant functions.

Okay I looked up an solution.

Let $f$ be an idempotent and define $$A=\{x | f(x)=1\},$$ $$B=\{x | f(x)=0\},$$ $f*f=f$ for any idempotent element, so $f=0$ or $f=1$ for any $x$ in $[0,1]$. So, $A \cup B= [0,1]$ and $A\cap B=\phi$. $A$ and $B$ are inverse image of a closed set by a continuous function so they are closed and bounded.

Let $c=\sup A$. It says we can assume $c \neq1$(wonder why...) then it is obvious that $f$ is not continuous at $c$(I wonder why). Any help please? I see $c$ must belong to $A$, since $A$ is closed...cannot proceed from there.

$\endgroup$
1
$\begingroup$

I'm not sure why they are proceeding in that fashion. A better way to go is this: since $A,B$ are disjoint and closed in $[0,1],$ and since $A\cup B=[0,1],$ then $A=[0,1]\setminus B$ is also open in $[0,1],$ and so is $B.$

Since $[0,1]$ is connected, then the only subsets of $[0,1]$ that are both closed and open are...what?

(Additionally, while you can conclude that $A,B$ are bounded by virtue of being subsets of $[0,1],$ this has nothing particular to do with being inverse images of closed sets.)

As for why we can assume that $c\ne 1$ in their approach, it's because if $\sup A=1,$ then $1\in A,$ so $1\notin B,$ and so instead we let $c=\sup B,$ whence $c\ne 1$ and we proceed as before. In particular, we should note that $(c,1]$ is a subset of (whichever set $c$ isn't the supremum of), but $c$ is not an element of that set, which contradicts the fact that both sets are closed. Alternately, we can consider $g=1-f,$ instead, which we can prove to be an idempotent whenever $f$ is, then let $A'=\{x:g(x)=1\},$ $B'=\{x:g(x)=0\}$ and note that $B=A',$ $A=B',$ so if $\sup A=1,$ then $c=\sup A'=\sup B<1,$ and so....

Regardless, it would have been better for the solution to have phrased it this way:

Note that $\sup A$ and $\sup B$ cannot both be equal to $1,$ for then $1$ would belong to both $A$ and $B$ by virtue of their being closed sets. On the other hand, one of $\sup A,\sup B$ must be equal to $1.$ Without loss of generality, then, we may assume that $\sup A<1.$ Letting $c=\sup A,$ we have....

$\endgroup$
  • $\begingroup$ actually this argument came to my mind just minutes ago...this is very clear and nice but maybe it is a too topological solution for an algebraic problem... i don't know haha. i mentioned inverse image to say A and B are closed, sorry to confuse you. thank you for your help! $\endgroup$ – Mathcho Sep 8 '15 at 14:47
  • 1
    $\begingroup$ You're very welcome! I have also expanded my answer to address what I suspect the solution's intent was. See what you think! $\endgroup$ – Cameron Buie Sep 8 '15 at 14:57
0
$\begingroup$

You aren't clear what the topology is, but from context it looks like you want the usual topology on [0,1]. If this is the case, the following solution seems easy:

You already know the range of f is the two element set {0,1} If the function is not constant, the ordinary Intermediate Value Theorem would tell us there is a value which f maps to 1/2, which is absurd.

Working with the preimages is instructive though. Thinking in this way, one can see how to connect idempotent continuous functions to pairs of disjoint clopen sets.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.