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## Math Q&A Game

 Quote by pwsnafu I just came across this: Let C be the Cantor set and $E = C \times [0,1]^{k-1} \subset R^k$. Then E is uncountable with cardinality c and with Lebesgue measure zero. So there are 2c subsets of E, each Lebesgue measurable.
This is close, but I am looking for subsets of the reals. So, set k to 1 and you have shown that lower limit is at least 2c. Now just provide an upper limit.

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 Quote by Jimmy Snyder This is close, but I am looking for subsets of the reals. So, set k to 1 and you have shown that lower limit is at least 2c. Now just provide an upper limit.
But isn't the cardinality of all subsets of the reals ##2^c##, so that is also an upper limit?

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 Quote by LCKurtz But isn't the cardinality of all subsets of the reals ##2^c##, so that is also an upper limit?
Yes, you have solved it.
 Topic: Two birds in the tree, the hunter shot one. Ask: Only a few were left in the tree? Live or die? You need to determine the answer

 Quote by godsaveme Only a few were left in the tree? Live or die?
What??

Sorry, I cannot parse those sentence fragments.

 Quote by DaveC426913 What?? Sorry, I cannot parse those sentence fragments.
I am a chinese,my english is poor.
That may be how many birds in the tree?
 This is a certainty and uncertainty question!
 How to determine? The number of birds，live or die？
 Can you help me to translate chinese into english?
 Zero. The other bird flew away. Edit: Or maybe one, if the bullet didn't knock the first bird off the perch. In any case, the live bird is gone.
 I don't think you can answer questions like that. There are two birds to begin with, one is shot dead. 1 is left. 1 does not equal a few, as a few is generally equated to mean 5.
 Actually I can answer questions like that. Proof: I just did I think the "a few" was just a translation failure on the part of our Chinese friend.
 Blog Entries: 1 I heard this one before. He: There were two birds in the yard and I shot one of them. How many were left in the yard? She: One. He: No, one. The one that I shot. The other one flew away.
 Okay, guess it's my turn to ask a new one. What are all the continuous functions $f:\mathbb{C} \rightarrow \mathbb{C}$ such that $\forall z,w\in \mathbb{C},\ f(z+w) = f(z)f(w)$? Does the answer change if continuous is replaced with measurable?

 Quote by jgutierrez218 ...a few is generally equated to mean 5.
What?? Where did you get this?

For me, "a few" is three or more.
 Blog Entries: 1 2 is a couple. 3 is a crowd. 3 to 7 is a few. 5 to 10 is some. 8 to 15 is several. 15 to 37 is a bunch or if it is something you don't like, then it's many, or even too many if you really don't like it. 30 - 100 is a profusion. 100 - 1000 is a multitude. More than that is a plethora or a surfeit.

 Quote by Jimmy Snyder 2 is a couple. 3 is a crowd. 3 to 7 is a few. 5 to 10 is some. 8 to 15 is several. 15 to 37 is a bunch or if it is something you don't like, then it's many, or even too many if you really don't like it. 30 - 100 is a profusion. 100 - 1000 is a multitude. More than that is a plethora or a surfeit.
Oh yes, so often do I ask for a crowd of things.

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