AS far as ideal of a ring is concerned, it is not ideal. I am giving an counter example. f(x) = 1 for all x belongs to R. which is a continuous function with compact support. g(x) = x for all x belongs to closed interval 0 to infinity.
= 0 otherwise
Now f(x) . g(x) has not a compact support.
definition of compact support – A function has compact support if it is zero outside of a compact set. Alternatively, one can say that a function has compact support if its support is a compact set.(what I know)
Can anyone please help me out ?
Best Answer
The set of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with compact support is indeed an ideal, in the ring-theoretic sense, in the ring of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$.
I think the source of your confusion is the definition of a compact set.
The function $f:\mathbb{R}\rightarrow\mathbb{R}: x\mapsto 1$ that you mention does not have compact support - its support (the closure of the set of all values on which it is nonzero) is $\mathbb{R}$, and this is not compact.
(The function $f$ is constant on a compact set, and its range is compact - but that's not the same as having compact support.)
Why is $\mathbb{R}$ not compact? Well, remember that a compact set is one with the property
But $\{(-n, n): n\in\mathbb{N}\}$ is a cover of $\mathbb{R}$ by open sets, with no finite subcover
Alternatively, in a Euclidean metric space (i.e., $\mathbb{R}^n$ for some $n$), a set is compact iff it is closed and bounded. Well, $\mathbb{R}$ is certainly closed, but is it bounded?
(Note that above I am considering $\mathbb{R}$ with the usual topology and metric, but since you didn't say otherwise I assume this is the context you are working in.)