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\newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}}
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Bolzan0Weierstrass Theorem

Every bounded sequence (\vec{x}_k)_{k=1}^{\infty} has a convergent subsequence.

Concepts

Did this not hold for multivariables? Something like that since a sequence can be in a circle (bounded) but not necessarly have a convergent subsequence? I gotta check again.

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon

Bolzan0Weierstrass Theorem

Every bounded sequence (\vec{x}_k)_{k=1}^{\infty} has a convergent subsequence.

Concepts

Did this not hold for multivariables? Something like that since a sequence can be in a circle (bounded) but not necessarly have a convergent subsequence? I gotta check again.

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof

Bolzan0Weierstrass Theorem

Every bounded sequence (\vec{x}_k)_{k=1}^{\infty} has a convergent subsequence.

Concepts

Did this not hold for multivariables? Something like that since a sequence can be in a circle (bounded) but not necessarly have a convergent subsequence? I gotta check again.

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon

Bolzan0Weierstrass Theorem

Every bounded sequence (\vec{x}_k)_{k=1}^{\infty} has a convergent subsequence.

Concepts

Did this not hold for multivariables? Something like that since a sequence can be in a circle (bounded) but not necessarly have a convergent subsequence? I gotta check again.

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof