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Although since the OP tag this as real analysis I assume he might be talking of any finite dimensional metric space, in which case the finite sets are always closed. $\endgroup$ – Alonso Delfín Commented Jun 8, 2015 at 21:56
Clearly every finite set is countable, but also some infinite sets are countable. Note that some places define countable as infinite and the above definition. In such cases we say that finite sets are "at most countable".
A word on the axiom of choice and finiteness.The above proof shows that finite sets are Dedekind-finite.There are other ways of defining finiteness, all which are true for finite sets, but may also be true for infinite sets.
A finite set is compact.Proof by induction. (1). If S = ∅ and F is any open family then S ⊂ ∪F and we may let G = ∅. (2). Suppose that n ≥ 0 and that every open cover of any n -member set has a finite subcover. Then if S = {x} ∪ T has n + 1 members with x ∉ T, and F is an open cover of S, then F is also an open cover of T, and T ...
Set A is finite iff every non-empty family of subsets of A has a minimal element [ordered by strict inclusion ' $\subset$ ']- A. Tarski via Suppes Furthermore, if one such family of subsets, F, has a minimal element , then it must also have a maximal element via the following argument:
Dec 26, 2015 at 20:27. In R^n define m to be the minimum among all pairwise distances between points; the existence of m relies on your finiteness assumption. Then an open ball of radius m around any point p in your set contains only p, which shows p is not a limit point. So there are no limit points in the set. – Benjamin Dickman.
A trivial example is every topology on a finite set: all subsets of a finite set are finite, thus also all compact subsets of this finite set are finite. Another example is the discrete topology (on any set), i.e. the topology where every one-element subset is open, or (equivalently) every subset is open.
0. From wikipedia: A set in a measure space X is said to have σ -finite measure if it is a countable union of sets with finite measure. Quick and easy question about the definition: Suppose a set E is σ -finite, and let {En}∞1 be a countable sequence of disjoint sets such that: ∞ ⋃ 1 En = E.
The union of two finite sets is a finite set. Let $X,Y$ be finite sets. That means that there are $n, m\in \omega$ such that $X \sim n$ and $Y \sim m$, i.e. there are ...
A "finite interval" is an interval of finite length, i.e, the number b − a b − a is finite. A finite set, on the other hand, is a set of finite cardinality, so consisting of only finitely many elements. Maybe the terminology is a bit unfortunate, but since EVERY nonempty interval possesses uncountably many elements, there is not much chance ...