$leq_{frak K_lambda}$-increasing continuous
$begingroup$
Here (in the context of Abstract Elementary Classes) on the page 43 at the bottom,-6th
line, what does it technically mean $$leq_{frak K_lambda}-text{increasing continuous}$$
? I think that this should be a condition on limit ordinals, but in his text, Shelah uses $alpha$ for both, limit and successors ordinals (see the page 67 in the link above) and he writes in that -6th line
[...for] $alpha<lambda^+$.
elementary-set-theory ordinals filtrations
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
$endgroup$
add a comment |
$begingroup$
Here (in the context of Abstract Elementary Classes) on the page 43 at the bottom,-6th
line, what does it technically mean $$leq_{frak K_lambda}-text{increasing continuous}$$
? I think that this should be a condition on limit ordinals, but in his text, Shelah uses $alpha$ for both, limit and successors ordinals (see the page 67 in the link above) and he writes in that -6th line
[...for] $alpha<lambda^+$.
elementary-set-theory ordinals filtrations
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
$endgroup$
add a comment |
$begingroup$
Here (in the context of Abstract Elementary Classes) on the page 43 at the bottom,-6th
line, what does it technically mean $$leq_{frak K_lambda}-text{increasing continuous}$$
? I think that this should be a condition on limit ordinals, but in his text, Shelah uses $alpha$ for both, limit and successors ordinals (see the page 67 in the link above) and he writes in that -6th line
[...for] $alpha<lambda^+$.
elementary-set-theory ordinals filtrations
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
$endgroup$
Here (in the context of Abstract Elementary Classes) on the page 43 at the bottom,-6th
line, what does it technically mean $$leq_{frak K_lambda}-text{increasing continuous}$$
? I think that this should be a condition on limit ordinals, but in his text, Shelah uses $alpha$ for both, limit and successors ordinals (see the page 67 in the link above) and he writes in that -6th line
[...for] $alpha<lambda^+$.
elementary-set-theory ordinals filtrations
elementary-set-theory ordinals filtrations
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
edited Dec 18 '18 at 5:14
Asaf Karagila♦
305k33435765
305k33435765
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
asked Dec 16 '18 at 18:22
user122424user122424
1,1232716
1,1232716
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note: the relevant definition is at the bottom of page $43$.
You aren't misunderstanding anything, but you are overthinking a bit. Shelah could have indeed written "for all limit $alpha<lambda^+$," as you observe, but he didn't need to: continuity is a vacuous condition at successor ordinals. Thinking about it topologically, each successor ordinal $beta$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $beta$.
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
add a comment |
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$begingroup$
Note: the relevant definition is at the bottom of page $43$.
You aren't misunderstanding anything, but you are overthinking a bit. Shelah could have indeed written "for all limit $alpha<lambda^+$," as you observe, but he didn't need to: continuity is a vacuous condition at successor ordinals. Thinking about it topologically, each successor ordinal $beta$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $beta$.
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
add a comment |
$begingroup$
Note: the relevant definition is at the bottom of page $43$.
You aren't misunderstanding anything, but you are overthinking a bit. Shelah could have indeed written "for all limit $alpha<lambda^+$," as you observe, but he didn't need to: continuity is a vacuous condition at successor ordinals. Thinking about it topologically, each successor ordinal $beta$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $beta$.
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
$endgroup$
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
add a comment |
$begingroup$
Note: the relevant definition is at the bottom of page $43$.
You aren't misunderstanding anything, but you are overthinking a bit. Shelah could have indeed written "for all limit $alpha<lambda^+$," as you observe, but he didn't need to: continuity is a vacuous condition at successor ordinals. Thinking about it topologically, each successor ordinal $beta$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $beta$.
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
$endgroup$
Note: the relevant definition is at the bottom of page $43$.
You aren't misunderstanding anything, but you are overthinking a bit. Shelah could have indeed written "for all limit $alpha<lambda^+$," as you observe, but he didn't need to: continuity is a vacuous condition at successor ordinals. Thinking about it topologically, each successor ordinal $beta$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $beta$.
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
answered Dec 16 '18 at 19:20
Noah SchweberNoah Schweber
125k10150287
125k10150287
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
add a comment |
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
You always respond kindly to my confusions. Thank you. Could you please also explain more closely the very last sentence from your answer? :each successor ordinal $β$ is an isolated point, and so there are no restrictions at all on how a continuous function needs to behave at $β$.
$endgroup$
– user122424
Dec 17 '18 at 15:35
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
@user122424 Think about how any map from a discrete space is continuous; this is a more general description of the same thing. If $X$ is a topological space, $ain X$ is an isolated point, and $f:Xrightarrow Y$ is a continuous map, then any $g:Xrightarrow Y$ which agrees with $f$ except possibly at $a$ is also continuous. Basically, isolated points are topologically uninteresting (at least, considered individually - the set of isolated points in a given space might actually be an interesting object). In particular, a function from an ordinal which is "continuous at limits" is continuous.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
$begingroup$
This is really only a useful comment if you're more familiar with point-set topology than set theory; if you're not, ignore it, it's only intended as intuitive motivation.
$endgroup$
– Noah Schweber
Dec 17 '18 at 16:02
add a comment |
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