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10h
awarded  elementary-set-theory
16h
answered Relationship between completeness and well ordering (meta).
17h
answered Problem on elementary logic and set theory
1d
answered How do I find the type of relation on an infinite set?
Jun
28
comment Weakening compactness in metric spaces
What is "the cardinaility of a net"? That of its domain or its image?
Jun
28
comment Can we define $ℝ^A$ where A is uncountable?
@AsafKaragila only if the domain is well-ordered I would use sequence.
Jun
28
answered Contraction on a metric.
Jun
20
answered Prove $((A^C \cup B^C) \setminus A)^C = A$
Jun
7
answered unclear why subset of a lexicographic ordered set of strings is said NOT to have a least element.
Jun
7
comment Generalization of metric spaces?
@StanCoreyCarter You're welcome!
Jun
7
answered Some basic set theory proofs.
Jun
7
revised Prove that sets being equipotent is an equivalence relation
tags, terminology, grammar
Jun
7
answered Generalization of metric spaces?
Jun
7
revised Prove that a separable metric space is Lindelöf without proving it is second-countable
added 2 characters in body
Jun
7
comment Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal element
The second case ($a_n$ does not proceed any $a_i$) needs some argument as you just claim the maximality of $a$ and $a_n$, not prove it.
Jun
7
comment Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal element
In the first case (some $a_i \ge a$), you rightly conclude $a = a_n$ which is a contradiction (as $a$ cannot be $a_n$ by definition), not a proof that $a$ is maximal. So you know $a$ does not proceed $a_n$ (as this led to the contradiction). Now you know $a$ is maximal by the void implication rgument in my answer.
Jun
6
comment Two sets $X,Y \subset [0,1]$ such that $X+Y=[0,2]$
There is an $\varepsilon$ in the exponent as well, so this makes it stronger than being measure $0$, maybe?
Jun
6
comment Uniform continuity on a dense subset implies uniform continuity on the set.
I think you are confused with another result about extending $f$ from $V_1$, here $f$ is already defined on all of $M_1$.
Jun
6
comment Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal element
Expand the argument and we'll see. Do you use the correct definition of maximal?
Jun
6
answered Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal element