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https://math.stackexchange.com/questions/3279942/topological-spaces-in-which-a-set-is-the-support-of-a-continuous-function-iff-it
An exercise of Rudin's Real and Complex Analysis says: Is it true that every compact subset of $\mathbf{R}^1$ is the support of a continuous function? If not, can you describe the class of all compact sets in $\mathbf{R}^1$ which are supports of continuous functions? Is your description valid in other topological spaces?
http://www.msc.uky.edu/ken/ma570/lectures/lecture2/html/compact.htm
The set of all the for all the is an open cover of and so it admits of a finite subcover . But then the set of all the 's associated with all the 's is a finite subcover of . Lemma 3: If every rectangle is compact, then every closed and bounded subset of is compact.
https://en.wikipedia.org/wiki/Support_(mathematics)
Every continuous function on a compact topological space has compact support since every closed subset of a compact space is indeed compact. Essential support [ edit ] If X is a topological measure space with a Borel measure μ (such as R n , or a Lebesgue measurable subset of R n , equipped with Lebesgue measure), then one typically identifies ...
https://math.stackexchange.com/questions/110573/continuous-mapping-on-a-compact-metric-space-is-uniformly-continuous
Continuous mapping on a compact metric space is uniformly continuous. Ask Question Asked 7 years, ... Are continuous functions on those sets uniformly continuous? Can you remember any theorems regarding those? ... Continuous with compact support implies uniform continuity. 2.
http://www.ams.org/journals/tran/1971-156-00/S0002-9947-1971-0275367-4/S0002-9947-1971-0275367-4.pdf
1. Introduction. The support of a real continuous function / on a topological space A" is the closure of the set of points in Afat which/does not vanish. Gillman and Jerison have shown that when A'is a realcompact space, the functions in C(X) with compact support are precisely the functions which belong to every free maximal ideal in C(X).
https://www.sciencedirect.com/science/article/pii/0001870865900411
Clearly, F is a closed linear subring with unit and contains SPACES OF CONTINUOUS FUNCTIONS ON COMPACT SETS 343 the complex conjugate of each of its elements, hence the set T(F) has the same properties and induces a continuous map from X onto the space of nonzero multiplicative linear functionals on F, which is homeomorphic to Z; we have only ...Cited by: 13
https://math.rice.edu/~semmes/fun5.pdf
5 Compact support 6 6 Inductive limits 8 7 Distributions 9 ... j=1 of continuous functions on U converges to a continuous function f on ... every compact set in Rn is contained in B(0,r) for some r ≥ 0, because compact subsets of Rn are bounded. This implies that one can get the same topology
http://www.math.jhu.edu/%7Efspinu/405/405-continuity%20thms.pdf
COMPACT SETS, CONNECTED SETS AND CONTINUOUS FUNCTIONS 1. Definitions 1.1. D ‰ Ris compact if and only if for any given open covering of D we can subtract a finite sucovering. That is, given (Gfi)fi 2 A a collection of open subsets of R(A an arbitrary set of indices)
https://www.quora.com/Is-there-a-non-compact-set-such-that-every-continuous-real-valued-function-on-the-set-attains-a-maximum-and-minimum
Consider the set of all real numbers with the cocountable topology. This means that the open sets are the empty set and every set whose complement is at most countable (call such a set cocountable). This is a topology: * the union of any number of...
https://en.wikipedia.org/wiki/Compact_space
Slightly more generally, this is true for an upper semicontinuous function.) As a sort of converse to the above statements, the pre-image of a compact space under a proper map is compact. Compact spaces and set operations. A closed subset of a compact space is compact, and a finite union of compact sets is compact.
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