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"We encourage children to read for enjoyment, yet we never encourage them to 'math' for enjoyment. We teach kids that math is done fast, done only one way and if you don't get the answer right, there's something wrong with you. You would never teach reading this way." - Rachel McAnallen

"Mathematicians create by acts of insight and intuition. Logic then sanctions the conquests of intuition." - Morris Kline


8h
comment Can every cardinal number between $\kappa^+$ and $2^\kappa$ be realized in this way?
@Rustyn, hehe thanks. I'm rather proud of it also :)
9h
revised Can every cardinal number between $\kappa^+$ and $2^\kappa$ be realized in this way?
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9h
asked Can every cardinal number between $\kappa^+$ and $2^\kappa$ be realized in this way?
21h
comment Order preserving functions that do not preserve binary operations
@Ebearr, no worries mate.
1d
comment When do Leibniz-like rules lead to unique linear operators
Neat question..
1d
accepted Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
1d
comment Order preserving functions that do not preserve binary operations
@Ebearr, I'm not sure I understand you. Meets, when they exist, are uniquely determined by the structure of the poset; you don't get to choose them. The meet of $x,y \in F$ is a special entity $x \wedge y \in F$ with the following important property: $a \leq x \wedge y$ iff $a \leq x$ and $a \leq y,$ for all $a \in F$. This determines $x \wedge y$ uniquely, if it exists. Now if $x \wedge y$ exists for all $x,y \in F$, like in the case you're dealing with, then this is equivalent to saying that $\wedge$ is a total function $F \times F \rightarrow F,$ in which it is necessarily associative.
1d
comment Order preserving functions that do not preserve binary operations
@Ebearr, pretty sure that if $f$ is a closure operator, then arbitrary meets in $\mathrm{Fix}(f)$ will coincide with intersections. Are you finding that's not the case? Also, meets are associative in any lattice.
1d
comment Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
@AsafKaragila, $\alpha$ is any ordinal. See my edit.
1d
revised Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
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1d
answered Order preserving functions that do not preserve binary operations
1d
revised Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
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1d
revised Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
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1d
revised Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
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1d
asked Cardinality: is it true that $|X^\mathcal{inj}| \leq 2^{|X|}$?
1d
comment Does distinct subformulas in this context mean syntactically distinct or semantically different?
Nice question. By the way, what are $A$ and $B$?
1d
revised Resolution in first order logic
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1d
answered Resolution in first order logic
Sep
28
comment Thinking and writing about mathematical structures in a way that is rigorous and precise.
@archipelago, fixed.
Sep
28
revised Thinking and writing about mathematical structures in a way that is rigorous and precise.
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