To prove that $ cap_n=1^infty (0 , frac1n) = emptyset $To show that $f(x)= frac1x$ is not uniformly continuous on $(0,infty)$Prove that $sigma$-algebra of subsets of $mathbbR$ of the form $(a,infty)$ contains all the intervals.Prove that $bigcap_n=1^infty$ $(0,frac1 n) = emptyset$Show that there is a unique $r$ in real separates $A$ and $B$, such that $(-infty, r)subset A$ and $(r,infty)subset B$.Prove that $bigcaplimits^infty_n=1 left(0, frac1n right)= emptyset$Disproving the proposition that $a_n = (-1)^nn$ has a limit.To prove $cap_n=1^infty A_n $ is non emptyA sequence of nested unbounded closed intervals $L_1supseteq L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n = varnothing$Let $a_n_n = 1^infty$ and $b_n_n = 1^infty$ be two sequences of real numbers s.t $|a_n -b_n| < frac1n$Using Archimedian property to prove that infimum of set is $0$
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To prove that $ cap_n=1^infty (0 , frac1n) = emptyset $
To show that $f(x)= frac1x$ is not uniformly continuous on $(0,infty)$Prove that $sigma$-algebra of subsets of $mathbbR$ of the form $(a,infty)$ contains all the intervals.Prove that $bigcap_n=1^infty$ $(0,frac1 n) = emptyset$Show that there is a unique $r$ in real separates $A$ and $B$, such that $(-infty, r)subset A$ and $(r,infty)subset B$.Prove that $bigcaplimits^infty_n=1 left(0, frac1n right)= emptyset$Disproving the proposition that $a_n = (-1)^nn$ has a limit.To prove $cap_n=1^infty A_n $ is non emptyA sequence of nested unbounded closed intervals $L_1supseteq L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n = varnothing$Let $a_n_n = 1^infty$ and $b_n_n = 1^infty$ be two sequences of real numbers s.t $|a_n -b_n| < frac1n$Using Archimedian property to prove that infimum of set is $0$
$begingroup$
$$ bigcap_n=1^infty left(0 , frac1nright) = varnothing$$
Now assume that intersection contains $b$. So, $ b < frac1n forall n in N$. Since $b > 0$, so we have by Archimedian property that $exists n in mathbbN$ such that $ frac1n < b$ which is a contradiction to assumption that $b < frac1n$
Is this correct ?
Thanks
real-analysis self-learning
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
$endgroup$
add a comment |
$begingroup$
$$ bigcap_n=1^infty left(0 , frac1nright) = varnothing$$
Now assume that intersection contains $b$. So, $ b < frac1n forall n in N$. Since $b > 0$, so we have by Archimedian property that $exists n in mathbbN$ such that $ frac1n < b$ which is a contradiction to assumption that $b < frac1n$
Is this correct ?
Thanks
real-analysis self-learning
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
$endgroup$
$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43
add a comment |
$begingroup$
$$ bigcap_n=1^infty left(0 , frac1nright) = varnothing$$
Now assume that intersection contains $b$. So, $ b < frac1n forall n in N$. Since $b > 0$, so we have by Archimedian property that $exists n in mathbbN$ such that $ frac1n < b$ which is a contradiction to assumption that $b < frac1n$
Is this correct ?
Thanks
real-analysis self-learning
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
$endgroup$
$$ bigcap_n=1^infty left(0 , frac1nright) = varnothing$$
Now assume that intersection contains $b$. So, $ b < frac1n forall n in N$. Since $b > 0$, so we have by Archimedian property that $exists n in mathbbN$ such that $ frac1n < b$ which is a contradiction to assumption that $b < frac1n$
Is this correct ?
Thanks
real-analysis self-learning
real-analysis self-learning
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
edited Mar 22 at 8:46
José Carlos Santos
173k23133241
173k23133241
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
asked Mar 22 at 8:40
J. DeffJ. Deff
717519
717519
$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43
add a comment |
$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43
$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43
$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Looks good to me.
Small notational nitpicks: It's Bbb N ("BlackBoard Bold") to make $Bbb N$.
And when using symbols to convey logic, like in your $b<frac1nforall nin Bbb N$, the quantifiers ($forall$ and $exists$) always come before whatever it is they modify.
So while one in English could say "$b$ is smaller than $frac1n$ for any natural number $n$", the symbolic statement must be $forall nin Bbb N, b<frac1n$ (exactly how to separate $forall nin Bbb N$ and $b<frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<frac1n$ in parentheses).
answered Mar 22 at 8:50
ArthurArthur
122k7122211
$endgroup$
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
oldest
votes
$begingroup$
Looks good to me.
Small notational nitpicks: It's Bbb N ("BlackBoard Bold") to make $Bbb N$.
And when using symbols to convey logic, like in your $b<frac1nforall nin Bbb N$, the quantifiers ($forall$ and $exists$) always come before whatever it is they modify.
So while one in English could say "$b$ is smaller than $frac1n$ for any natural number $n$", the symbolic statement must be $forall nin Bbb N, b<frac1n$ (exactly how to separate $forall nin Bbb N$ and $b<frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<frac1n$ in parentheses).
answered Mar 22 at 8:50
ArthurArthur
122k7122211
$endgroup$
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
add a comment |
$begingroup$
Looks good to me.
Small notational nitpicks: It's Bbb N ("BlackBoard Bold") to make $Bbb N$.
And when using symbols to convey logic, like in your $b<frac1nforall nin Bbb N$, the quantifiers ($forall$ and $exists$) always come before whatever it is they modify.
So while one in English could say "$b$ is smaller than $frac1n$ for any natural number $n$", the symbolic statement must be $forall nin Bbb N, b<frac1n$ (exactly how to separate $forall nin Bbb N$ and $b<frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<frac1n$ in parentheses).
answered Mar 22 at 8:50
ArthurArthur
122k7122211
$endgroup$
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
add a comment |
$begingroup$
Looks good to me.
Small notational nitpicks: It's Bbb N ("BlackBoard Bold") to make $Bbb N$.
And when using symbols to convey logic, like in your $b<frac1nforall nin Bbb N$, the quantifiers ($forall$ and $exists$) always come before whatever it is they modify.
So while one in English could say "$b$ is smaller than $frac1n$ for any natural number $n$", the symbolic statement must be $forall nin Bbb N, b<frac1n$ (exactly how to separate $forall nin Bbb N$ and $b<frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<frac1n$ in parentheses).
answered Mar 22 at 8:50
ArthurArthur
122k7122211
$endgroup$
Looks good to me.
Small notational nitpicks: It's Bbb N ("BlackBoard Bold") to make $Bbb N$.
And when using symbols to convey logic, like in your $b<frac1nforall nin Bbb N$, the quantifiers ($forall$ and $exists$) always come before whatever it is they modify.
So while one in English could say "$b$ is smaller than $frac1n$ for any natural number $n$", the symbolic statement must be $forall nin Bbb N, b<frac1n$ (exactly how to separate $forall nin Bbb N$ and $b<frac1n$ is up to you, you can use a comma like I did, or a colon, or wrap $b<frac1n$ in parentheses).
answered Mar 22 at 8:50
ArthurArthur
122k7122211
answered Mar 22 at 8:50
ArthurArthur
122k7122211
answered Mar 22 at 8:50
ArthurArthur
122k7122211
answered Mar 22 at 8:50
ArthurArthur
122k7122211
122k7122211
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
add a comment |
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
$begingroup$
"..thus we have $f$ continuous for every $xin D$.." opposed to "..thus for every $xin D$ $f$ is continuous.." or do you regard worded and symbolised quantifications differently?
$endgroup$
– Alvin Lepik
Mar 22 at 8:53
2
2
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
$begingroup$
@AlvinLepik I definitely regard worded and symbolic quantifiers differently. When using regular, human English, putting quantifiers after is natural, because that's how English semantics work.
$endgroup$
– Arthur
Mar 22 at 8:54
add a comment |
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$begingroup$
That's right. The same reasoning I used for a different problem.
$endgroup$
– Balakrishnan Rajan
Mar 22 at 8:43