# Describe the set of all compact sets which are supports of continuous functions

1. Nov 8, 2005

### benorin

The question reads: Is it true that every compact subset of $$\mathbb{R}$$ is the support of a continuous function? If not, can you describe the class of all compact sets in $$\mathbb{R}$$ which are supports of continuous functions? Is your description valid in other topological spaces?

The answer to the first question is no. A singleton is compact but is not the support of any continuous function; the same is true of the Cantor set (for it contains no segment). I am struck on the second question.

A continuous function is (in our text) defined as a function $$f:X\rightarrow Y$$ for topological spaces X and Y such that $$f^{-1}\left( V\right)$$ is an open set in X for every open set V in Y.

The support of a function is the closure of the set set of all values at which it is not zero, that is $$\overline{\left\{ x:f(x) \mbox{ not }= 0\right\} }$$

So I need to describe

$$\left\{ K\subset \mathbb{R}: \exists \mbox{ a continuous function } f \mbox{ such that support}(f)=K\right\} \cap \left\{ K\subset \mathbb{R}:K \mbox{ is compact} \right\}$$

Based on the Cantor set example given as a counter-example to the first question, I'm guessing that connectedness may be involved, but I really don't know. How can I answer this question so that it holds for a general topological space?

-Ben

Last edited: Nov 8, 2005
2. Nov 8, 2005

### Hurkyl

Staff Emeritus
My gut says that you should look at the complementary condition -- look at where the functions are zero, and see what that says about the support.

As for a general topological space, . They're too yucky! So, I suspect that whatever you say for the second question, there's a topological space that contains a counterexample. I would not be surprised if there isn't an easy condition that holds in the general case.

Last edited: Nov 8, 2005
3. Nov 8, 2005

### AKG

Your definition of "support" says something about f(x) = 0, so how does this problem extend to general topological spaces, i.e. what if f : X -> Y is such that Y does not contain 0? Anyways, let's just deal with f : R -> R for now.

support(f) = closure(f-1(R - {0}))

So it seems we want to know: in what cases is the thing on the right closed and bounded? Well it is always closed, so when is it bounded? It is bounded iff f-1(R-{0}) is bounded. It seems to me that a compact set will be a support iff it is the closure of an open set. The Cantor set is not the closure of an open set, nor is a singleton.

So claim: A compact set is a support iff it is the closure of an open set.

Proof, if it is not the closure of an open set, then it is not, in particular, the closure of f-1(R-0) which is an open set since f is continuous and R-0 is open in R, and thus it is not the support of f. Conversely, if it is the closure of a bounded open set, U, then can we define a continuous function f which is non-zero on U and zero elsewhere? If so, then certainly it is a support.

The answer to the bold question certainly seems plausible. An open set is a union of open intervals (since the open intervals form a basis). We can choose to consider U as a union of disjoint open intervals. We can define functions on these disjoint intervals which are non-zero on them, and zero outside of them, and then choose our f to be the sum of all these functions. So our problem reduces to finding continuous functions on a single open interval that is non-zero on the interval and vanishes identically outside the interval. Not only can we do this, but we can find Cinfinity functions that do this. The construction of such a function involves a bunch of exponential functions. You can generalize from the following example for the interval (-1,1):

$$f(x) = \exp \left[-(x-1)^{-2}\right ]\exp \left [-(x+1)^{-2}\right ]\ \mbox{for}\ x \in (-1,1),\ \ f(x) = 0\ \mbox{otherwise}$$

4. Nov 12, 2005

### benorin

Clairification on definition of support of a function

The support of a complex function $$f:X\rightarrow\mathbb{C}$$ on a topological space X is the closure of the set [parenthetical question: Hey, what's TeX for "not equal" ?]: $\left\{ x:f(x) \mbox{ not }= 0\right\}$.

My bad, I had generalized from $$\mathbb{C}$$ to any topological space Y in a previous post: that is no generalization at all. It doesn't even make sense :yuck: .

The generalization refered to in the question is that of answering the prior question for $$f:X\rightarrow\mathbb{C}$$ in stead of for $$f:\mathbb{R}\rightarrow\mathbb{C}$$.

Last edited: Nov 12, 2005
5. Nov 12, 2005

### Oxymoron

The TeX for not equal is "\neq" which can be seen here

$$1 \neq 0$$

6. Nov 12, 2005

### benorin

$\neq$ isn't in the LaTeX code reference PDF oft linked to. shame.

7. Nov 12, 2005

### mathwonk

the discussion seems to suggest looking at the interior of the sets considered.