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seen Jul 31 '11 at 18:34
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Jul
31
comment Proving that $(b_n) \to b$ implies $\left(\frac{1}{b_n}\right) \to \frac{1}{b}$
This makes sense, thank you. You are right, picking a smaller epsilon actually leads me closer to the sequence, I missed that at first. Thanks again!
Jul
31
accepted Proving that $(b_n) \to b$ implies $\left(\frac{1}{b_n}\right) \to \frac{1}{b}$
Jul
31
revised Proving that $(b_n) \to b$ implies $\left(\frac{1}{b_n}\right) \to \frac{1}{b}$
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Jul
31
asked Proving that $(b_n) \to b$ implies $\left(\frac{1}{b_n}\right) \to \frac{1}{b}$
Jul
28
awarded  Nice Question
Jul
12
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
@Asaf: Thanks for the link and for your answer. I have been running into the issue of not knowing which answer to accept on numerous occasions now (I feel like more than half of them deserve being accepted, if not all), and I certainly haven't been giving them that much time... I will do better next time.
Jul
12
accepted What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
Jul
12
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
Bruno, since I can't use cardinal arithmetic, I don't think I am allowed to make the step of $|\{\text{functions } \{0,1\} \to \mathbb{N} \}|= |\mathbb{N}^{\{0,1\}}|= |\mathbb{N}^2|$, I think I just don't "know" this is true. OHH nevermind, I see your update, I think it makes sense now! Thanks
Jul
12
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
Thanks, but I don't think I can use any cardinal arithmetic, just general wit... I have proved than $\mathbb{N}^2$ is equinumerous to $\mathbb{N}$ before, it's a question of getting to $\mathbb{N}^2$ w/o cardinal arithmetic here, I guess.
Jul
12
asked What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
Jul
11
accepted Proving the countability of algebraic numbers
Jul
10
asked Proving the countability of algebraic numbers
Jul
8
comment How to show that $\sqrt{2}+\sqrt{3}$ is algebraic?
An amazing answer Amitesh, thank you. One question, what do you mean when you say "Over Q?" I don't know what people mean when in math they say "bla bla bla over R/Q/etc." especially when talking about fields. Does it mean we take the coefficients of the polynomials be only rational numbers? Thanks!
Jul
8
accepted How to show that $\sqrt{2}+\sqrt{3}$ is algebraic?
Jul
8
revised How to show that $\sqrt{2}+\sqrt{3}$ is algebraic?
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