Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Nov 8, 2005 #1


    User Avatar
    Homework Helper

    The question reads: Is it true that every compact subset of [tex]\mathbb{R}[/tex] is the support of a continuous function? If not, can you describe the class of all compact sets in [tex]\mathbb{R}[/tex] 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 [tex]f:X\rightarrow Y[/tex] for topological spaces X and Y such that [tex]f^{-1}\left( V\right)[/tex] 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 [tex]\overline{\left\{ x:f(x) \mbox{ not }= 0\right\} }[/tex]

    So I need to describe

    [tex]\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\}[/tex]

    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?

    Please help.

    Last edited: Nov 8, 2005
  2. jcsd
  3. Nov 8, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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, :cry:. They're too yucky! So, I suspect that whatever you say for the second question, there's a topological space that contains a counterexample. :smile: I would not be surprised if there isn't an easy condition that holds in the general case.
    Last edited: Nov 8, 2005
  4. Nov 8, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    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):

    [tex]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}[/tex]
  5. Nov 12, 2005 #4


    User Avatar
    Homework Helper

    Clairification on definition of support of a function

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

    My bad, I had generalized from [tex]\mathbb{C}[/tex] 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 [tex]f:X\rightarrow\mathbb{C}[/tex] in stead of for [tex]f:\mathbb{R}\rightarrow\mathbb{C}[/tex].
    Last edited: Nov 12, 2005
  6. Nov 12, 2005 #5
    The TeX for not equal is "\neq" which can be seen here

    [tex]1 \neq 0[/tex]
  7. Nov 12, 2005 #6


    User Avatar
    Homework Helper

    [itex]\neq[/itex] isn't in the LaTeX code reference PDF oft linked to. shame.
  8. Nov 12, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    the discussion seems to suggest looking at the interior of the sets considered.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook