Nov 18, 2012 · 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. For …

Dec 15, 2014 · $ [0,1]$ is certainly not finite. Every finite set is compact, but the converse is false: take any infinite set with the indiscrete topology for an easy example.

Nov 2, 2013 · A (non-empty) finite totally ordered set has a maximum and minimum as in the proof of the accepted answer because of transitivity. A well-known set with a non-transitive …

3 Halmos' Naive Set Theory gives a hint which is the following Lemma. It took me awhile to understand the hint's usefulness for proving that the union of two finite sets is finite. The …

3 The parenthetical remark is just saying that formally the definition of finite set does not apply to the empty set, but the empty set is taken to be finite by convention. If this bothers you, note it …

The length of the edges will define your metric which is a scaling of the standard discrete metric iff you have an equilateral triangle. Of course, they all generate the same discrete topology on …

Jun 9, 2015 · So, both of these are open). Also, we know that the finite union of closed sets is closed and hence every finite set is closed in a metric space.

In set theory, you define $0 = \emptyset$, $1 = \ {\emptyset\}$, $2 = \ {\emptyset, \ {\emptyset\}\}$. In general ordinals (which include the finite ones), can be thought of as the set of ordinal …

Nov 12, 2021 · Take away: If a set is finite, then it has an infimum which is the minimum of that set, and a supremum which is the maximum of that set. Now for infinite sets, the minimum, …

Mar 24, 2023 · Proof that subtracting a finite set from an infinite one retains cardinality Ask Question Asked 2 years, 8 months ago Modified 2 years, 8 months ago