Work on Countable bouquet of circles (S139) #1601
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spaces/S000139/properties/P000214.md
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In my opinion it is easy enough to not argue why this doesn't converge, but if needed I can give a more precise argument (is a bit messy though)
spaces/S000139/properties/P000214.md
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I think instead you should justify ¬P214 for S131, and then use hereditariness, like what P63 does.
(Remark: It also applies for S135)
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I looked at this, and using S131 (or a similar argument) the alpha property follows for a few spaces. However are you sure about S135?
If I understand this correctly, it seems to me like the only sequences converging to 0 are eventually constant
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(Side remark: The picture is misleading if it is trying to suggest a nbhd of the origin. That does not match the text at all.)
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Fix a direction (line), the sequence that converge to 0 on that line converge to 0 on S135.
In fact, S131 is a closed subspace of S135.
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I dont really understand the definition I think:
Why do we need the second case in the definition (and instead remove the disjoint condition in the first case)? What's the difference
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of S&S's space 141 (in the picture)
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Because if the set contains 0, it must have every direction inhabited.
If does not contain 0, we don't need this condition.
Another equivalent definition: continuum many lines, each pick one point, and take quotient.
(From this definition we can see it has a strong relation with S139)
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Because if the set contains 0, it must have every direction inhabited. If does not contain 0, we don't need this condition.
Another equivalent definition: continuum many lines, each pick one point, and take quotient.
Oh I see, thank you. I overlooked the union I guess. :)
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I deleted P214 for now, since if we use the argument with S131, this applies more broadly to a couple of other spaces (131, 207, 135, 107, 97 and maybe some others), which would probably be preferable in a seperate PR |
Here we go again. #1600