Proving an infinite set not having any infinite decidable subsets is immune
Or, in other words, prove that an infinite set $S subset mathbb{N}$ that has no infinite decidable subsets also has no infinite enumerable subsets.
One idea I had is to show that the complement of $S$ has a non-empty intersection with any infinite enumerable set (kind of the reverse of how the existence of simple sets is proven), but since $S$ is not something we construct, I'm not sure how to proceed.
logic computability
asked Nov 25 at 20:55
0xd34df00d
397212
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Or, in other words, prove that an infinite set $S subset mathbb{N}$ that has no infinite decidable subsets also has no infinite enumerable subsets.
One idea I had is to show that the complement of $S$ has a non-empty intersection with any infinite enumerable set (kind of the reverse of how the existence of simple sets is proven), but since $S$ is not something we construct, I'm not sure how to proceed.
logic computability
asked Nov 25 at 20:55
0xd34df00d
397212
add a comment |
Or, in other words, prove that an infinite set $S subset mathbb{N}$ that has no infinite decidable subsets also has no infinite enumerable subsets.
One idea I had is to show that the complement of $S$ has a non-empty intersection with any infinite enumerable set (kind of the reverse of how the existence of simple sets is proven), but since $S$ is not something we construct, I'm not sure how to proceed.
logic computability
asked Nov 25 at 20:55
0xd34df00d
397212
Or, in other words, prove that an infinite set $S subset mathbb{N}$ that has no infinite decidable subsets also has no infinite enumerable subsets.
One idea I had is to show that the complement of $S$ has a non-empty intersection with any infinite enumerable set (kind of the reverse of how the existence of simple sets is proven), but since $S$ is not something we construct, I'm not sure how to proceed.
logic computability
logic computability
asked Nov 25 at 20:55
0xd34df00d
397212
asked Nov 25 at 20:55
0xd34df00d
397212
asked Nov 25 at 20:55
0xd34df00d
397212
asked Nov 25 at 20:55
0xd34df00d
397212
asked Nov 25 at 20:55
0xd34df00d
397212
397212
add a comment |
add a comment |
1 Answer
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Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.
answered Nov 26 at 18:28
0xd34df00d
397212
add a comment |
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Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.
answered Nov 26 at 18:28
0xd34df00d
397212
add a comment |
Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.
answered Nov 26 at 18:28
0xd34df00d
397212
add a comment |
Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.
answered Nov 26 at 18:28
0xd34df00d
397212
Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.
answered Nov 26 at 18:28
0xd34df00d
397212
answered Nov 26 at 18:28
0xd34df00d
397212
answered Nov 26 at 18:28
0xd34df00d
397212
answered Nov 26 at 18:28
0xd34df00d
397212
397212
add a comment |
add a comment |
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