Mixing
Another Proof: Bolzano Weierstrass Theorem
$begingroup$
I know that there are other posts discussing proofs of the BW Theorem. However, I could not find a post that discussed my specific proof. So, I have made a new post.
If the proof is correct, but lacks refinement, please allow me your feedback. It will be greatly appreciated, as I'm not only trying to improve my understanding of math, but also my communication of it.
Let $(a_k)$ be a bounded sequence, and define the set
$$S = {x in mathbb{R}: x < a_k textrm{ for infinitely many $a_k$}}.$$
Given that $S$ is bounded, as $(a_k)$ is bounded, the set $S$ has a supremum. I will show that there exists a subsequence $(a_{k_i})$ of $(a_k)$ that converges to $operatorname{sup}S,$ hence demonstrating that there exists a convergent subsequence of $(a_k).$
Note that for all $n in mathbb{N},$ there may be only finitely many elements $a_{k}$ in $(a_k)$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_{k}.$$ If there were some $n in mathbb{N}$ for which the aforementioned inequality holds for infinitely many elements $a_{k}$ in $(a_k),$ the set $S$ would contain elements $s in [operatorname{sup}S, operatorname{sup}S+frac{1}{n}),$ contradicting that $operatorname{sup}S$ is the supremum of $S.$
Also, note that for all $n in mathbb{N},$ there exists $s in S$ such that $$operatorname{sup}S - frac{1}{n} < s < operatorname{sup}S,$$ which implies that for all $n in mathbb{N},$ there exist infinitely many elements $a_k$ in $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{n} < s < a_k.$$
Given that for $n in mathbb{N}$ there exists only finitely many elements $a_k$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_k,$$ it follows that there for all $n in mathbb{N}$ there exist infinitely many elements $a_k$ that satisfy $$operatorname{sup}S - frac{1}{n} < a_k leq operatorname{sup}S + frac{1}{n}.$$
A subsequence $(a_{k_i})$ that converges to $operatorname{sup}S$ has now become clear. Given $i in mathbb{N},$ there exists an element $a_{k_i}$ such that $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}.$$ As $k_i$ is finite, there are only finitely many elements that appear before $a_{k_i}$ in the sequence $(a_k).$ As there are infinitely many terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i+1} < a_k leq operatorname{sup}S + frac{1}{i+1},$$ there exists $a_{k_{i+1}}$ such that $$operatorname{sup}S - frac{1}{i+1} < a_{k_{i+1}} leq operatorname{sup}S + frac{1}{i+1} quad textrm{and} quad k_{i+1} > k_{i}.$$
Hence, a convergent subsequence $(a_{k_i})$ may be constructed by choosing inductively terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}$$ for $i = 1, 2, ...$
real-analysis sequences-and-series proof-verification
edited Feb 26 at 9:24
user21820
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
add a comment |
$begingroup$
I know that there are other posts discussing proofs of the BW Theorem. However, I could not find a post that discussed my specific proof. So, I have made a new post.
If the proof is correct, but lacks refinement, please allow me your feedback. It will be greatly appreciated, as I'm not only trying to improve my understanding of math, but also my communication of it.
Let $(a_k)$ be a bounded sequence, and define the set
$$S = {x in mathbb{R}: x < a_k textrm{ for infinitely many $a_k$}}.$$
Given that $S$ is bounded, as $(a_k)$ is bounded, the set $S$ has a supremum. I will show that there exists a subsequence $(a_{k_i})$ of $(a_k)$ that converges to $operatorname{sup}S,$ hence demonstrating that there exists a convergent subsequence of $(a_k).$
Note that for all $n in mathbb{N},$ there may be only finitely many elements $a_{k}$ in $(a_k)$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_{k}.$$ If there were some $n in mathbb{N}$ for which the aforementioned inequality holds for infinitely many elements $a_{k}$ in $(a_k),$ the set $S$ would contain elements $s in [operatorname{sup}S, operatorname{sup}S+frac{1}{n}),$ contradicting that $operatorname{sup}S$ is the supremum of $S.$
Also, note that for all $n in mathbb{N},$ there exists $s in S$ such that $$operatorname{sup}S - frac{1}{n} < s < operatorname{sup}S,$$ which implies that for all $n in mathbb{N},$ there exist infinitely many elements $a_k$ in $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{n} < s < a_k.$$
Given that for $n in mathbb{N}$ there exists only finitely many elements $a_k$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_k,$$ it follows that there for all $n in mathbb{N}$ there exist infinitely many elements $a_k$ that satisfy $$operatorname{sup}S - frac{1}{n} < a_k leq operatorname{sup}S + frac{1}{n}.$$
A subsequence $(a_{k_i})$ that converges to $operatorname{sup}S$ has now become clear. Given $i in mathbb{N},$ there exists an element $a_{k_i}$ such that $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}.$$ As $k_i$ is finite, there are only finitely many elements that appear before $a_{k_i}$ in the sequence $(a_k).$ As there are infinitely many terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i+1} < a_k leq operatorname{sup}S + frac{1}{i+1},$$ there exists $a_{k_{i+1}}$ such that $$operatorname{sup}S - frac{1}{i+1} < a_{k_{i+1}} leq operatorname{sup}S + frac{1}{i+1} quad textrm{and} quad k_{i+1} > k_{i}.$$
Hence, a convergent subsequence $(a_{k_i})$ may be constructed by choosing inductively terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}$$ for $i = 1, 2, ...$
real-analysis sequences-and-series proof-verification
edited Feb 26 at 9:24
user21820
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36
add a comment |
$begingroup$
I know that there are other posts discussing proofs of the BW Theorem. However, I could not find a post that discussed my specific proof. So, I have made a new post.
If the proof is correct, but lacks refinement, please allow me your feedback. It will be greatly appreciated, as I'm not only trying to improve my understanding of math, but also my communication of it.
Let $(a_k)$ be a bounded sequence, and define the set
$$S = {x in mathbb{R}: x < a_k textrm{ for infinitely many $a_k$}}.$$
Given that $S$ is bounded, as $(a_k)$ is bounded, the set $S$ has a supremum. I will show that there exists a subsequence $(a_{k_i})$ of $(a_k)$ that converges to $operatorname{sup}S,$ hence demonstrating that there exists a convergent subsequence of $(a_k).$
Note that for all $n in mathbb{N},$ there may be only finitely many elements $a_{k}$ in $(a_k)$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_{k}.$$ If there were some $n in mathbb{N}$ for which the aforementioned inequality holds for infinitely many elements $a_{k}$ in $(a_k),$ the set $S$ would contain elements $s in [operatorname{sup}S, operatorname{sup}S+frac{1}{n}),$ contradicting that $operatorname{sup}S$ is the supremum of $S.$
Also, note that for all $n in mathbb{N},$ there exists $s in S$ such that $$operatorname{sup}S - frac{1}{n} < s < operatorname{sup}S,$$ which implies that for all $n in mathbb{N},$ there exist infinitely many elements $a_k$ in $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{n} < s < a_k.$$
Given that for $n in mathbb{N}$ there exists only finitely many elements $a_k$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_k,$$ it follows that there for all $n in mathbb{N}$ there exist infinitely many elements $a_k$ that satisfy $$operatorname{sup}S - frac{1}{n} < a_k leq operatorname{sup}S + frac{1}{n}.$$
A subsequence $(a_{k_i})$ that converges to $operatorname{sup}S$ has now become clear. Given $i in mathbb{N},$ there exists an element $a_{k_i}$ such that $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}.$$ As $k_i$ is finite, there are only finitely many elements that appear before $a_{k_i}$ in the sequence $(a_k).$ As there are infinitely many terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i+1} < a_k leq operatorname{sup}S + frac{1}{i+1},$$ there exists $a_{k_{i+1}}$ such that $$operatorname{sup}S - frac{1}{i+1} < a_{k_{i+1}} leq operatorname{sup}S + frac{1}{i+1} quad textrm{and} quad k_{i+1} > k_{i}.$$
Hence, a convergent subsequence $(a_{k_i})$ may be constructed by choosing inductively terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}$$ for $i = 1, 2, ...$
real-analysis sequences-and-series proof-verification
edited Feb 26 at 9:24
user21820
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
I know that there are other posts discussing proofs of the BW Theorem. However, I could not find a post that discussed my specific proof. So, I have made a new post.
If the proof is correct, but lacks refinement, please allow me your feedback. It will be greatly appreciated, as I'm not only trying to improve my understanding of math, but also my communication of it.
Let $(a_k)$ be a bounded sequence, and define the set
$$S = {x in mathbb{R}: x < a_k textrm{ for infinitely many $a_k$}}.$$
Given that $S$ is bounded, as $(a_k)$ is bounded, the set $S$ has a supremum. I will show that there exists a subsequence $(a_{k_i})$ of $(a_k)$ that converges to $operatorname{sup}S,$ hence demonstrating that there exists a convergent subsequence of $(a_k).$
Note that for all $n in mathbb{N},$ there may be only finitely many elements $a_{k}$ in $(a_k)$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_{k}.$$ If there were some $n in mathbb{N}$ for which the aforementioned inequality holds for infinitely many elements $a_{k}$ in $(a_k),$ the set $S$ would contain elements $s in [operatorname{sup}S, operatorname{sup}S+frac{1}{n}),$ contradicting that $operatorname{sup}S$ is the supremum of $S.$
Also, note that for all $n in mathbb{N},$ there exists $s in S$ such that $$operatorname{sup}S - frac{1}{n} < s < operatorname{sup}S,$$ which implies that for all $n in mathbb{N},$ there exist infinitely many elements $a_k$ in $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{n} < s < a_k.$$
Given that for $n in mathbb{N}$ there exists only finitely many elements $a_k$ that satisfy $$operatorname{sup}S + frac{1}{n} leq a_k,$$ it follows that there for all $n in mathbb{N}$ there exist infinitely many elements $a_k$ that satisfy $$operatorname{sup}S - frac{1}{n} < a_k leq operatorname{sup}S + frac{1}{n}.$$
A subsequence $(a_{k_i})$ that converges to $operatorname{sup}S$ has now become clear. Given $i in mathbb{N},$ there exists an element $a_{k_i}$ such that $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}.$$ As $k_i$ is finite, there are only finitely many elements that appear before $a_{k_i}$ in the sequence $(a_k).$ As there are infinitely many terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i+1} < a_k leq operatorname{sup}S + frac{1}{i+1},$$ there exists $a_{k_{i+1}}$ such that $$operatorname{sup}S - frac{1}{i+1} < a_{k_{i+1}} leq operatorname{sup}S + frac{1}{i+1} quad textrm{and} quad k_{i+1} > k_{i}.$$
Hence, a convergent subsequence $(a_{k_i})$ may be constructed by choosing inductively terms in the sequence $(a_k)$ that satisfy $$operatorname{sup}S - frac{1}{i} < a_{k_i} leq operatorname{sup}S + frac{1}{i}$$ for $i = 1, 2, ...$
real-analysis sequences-and-series proof-verification
real-analysis sequences-and-series proof-verification
edited Feb 26 at 9:24
user21820
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
edited Feb 26 at 9:24
user21820
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
edited Feb 26 at 9:24
user21820
39.8k544158
edited Feb 26 at 9:24
user21820
39.8k544158
edited Feb 26 at 9:24
user21820
39.8k544158
39.8k544158
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
asked Jan 28 at 22:38
Rafael VergnaudRafael Vergnaud
357217
357217
$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36
add a comment |
$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36
$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36
add a comment |
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$begingroup$
You just need to reduce your proof in length. Your argument begins from "Given $iinmathbb {N}$..." and you can remove everything before that especially those dealing with $ninmathbb {N}$.
$endgroup$
– Paramanand Singh
Jan 29 at 8:25
$begingroup$
Hey, Singh. Thank you for the feedback. I fell as if the part before that was necessary to explain why what came next is true. No?
$endgroup$
– Rafael Vergnaud
Jan 30 at 3:57
$begingroup$
To me it looks like repetition, but others may think differently. Some explanation can be given, but using two different symbols $i, n$ for the same thing can be confusing.
$endgroup$
– Paramanand Singh
Jan 30 at 4:36