Mixing
Prove a double inclusion of an arbitrary unions
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$B_n$ is defined as follows:
$$B_n = ; 3n+1 < x le 3n+4 $$
What I need to do is:
- "Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$:
$$bigcup_1 le n in mathbbNB_n$$ - Prove the calculation using a double inclusion (showing that each set is a subset of the other).
What I currently have:
- Considering that $1 le n$, I "calculated" the arbitrary unions of $B_n$ to be:
$$bigcup_1 le n in mathbbNB_n = x in mathbbN ;$$ Proving the first direction:
Let:
$$ x in bigcup_1 le n in mathbbNB_n $$Since $1 le n in mathbbN$, then also $4 < x in mathbbN$.
Thus:
$$ bigcup_1 le n in mathbbNB_n subseteq x in mathbbN ; $$
Now, I'm not sure if my first-direction proof is correctly written - and please correct me if it's not - but what's more of an issue here is this:
How do I approach the opposite direction proof?
I'd be glad for any guidance.
elementary-set-theory proof-writing
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
$B_n$ is defined as follows:
$$B_n = ; 3n+1 < x le 3n+4 $$
What I need to do is:
- "Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$:
$$bigcup_1 le n in mathbbNB_n$$ - Prove the calculation using a double inclusion (showing that each set is a subset of the other).
What I currently have:
- Considering that $1 le n$, I "calculated" the arbitrary unions of $B_n$ to be:
$$bigcup_1 le n in mathbbNB_n = x in mathbbN ;$$ Proving the first direction:
Let:
$$ x in bigcup_1 le n in mathbbNB_n $$Since $1 le n in mathbbN$, then also $4 < x in mathbbN$.
Thus:
$$ bigcup_1 le n in mathbbNB_n subseteq x in mathbbN ; $$
Now, I'm not sure if my first-direction proof is correctly written - and please correct me if it's not - but what's more of an issue here is this:
How do I approach the opposite direction proof?
I'd be glad for any guidance.
elementary-set-theory proof-writing
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$B_n$ is defined as follows:
$$B_n = ; 3n+1 < x le 3n+4 $$
What I need to do is:
- "Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$:
$$bigcup_1 le n in mathbbNB_n$$ - Prove the calculation using a double inclusion (showing that each set is a subset of the other).
What I currently have:
- Considering that $1 le n$, I "calculated" the arbitrary unions of $B_n$ to be:
$$bigcup_1 le n in mathbbNB_n = x in mathbbN ;$$ Proving the first direction:
Let:
$$ x in bigcup_1 le n in mathbbNB_n $$Since $1 le n in mathbbN$, then also $4 < x in mathbbN$.
Thus:
$$ bigcup_1 le n in mathbbNB_n subseteq x in mathbbN ; $$
Now, I'm not sure if my first-direction proof is correctly written - and please correct me if it's not - but what's more of an issue here is this:
How do I approach the opposite direction proof?
I'd be glad for any guidance.
elementary-set-theory proof-writing
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$B_n$ is defined as follows:
$$B_n = ; 3n+1 < x le 3n+4 $$
What I need to do is:
- "Calculate" (not sure if that's the correct term, please correct me with the right one) the arbitrary unions of $B_n$:
$$bigcup_1 le n in mathbbNB_n$$ - Prove the calculation using a double inclusion (showing that each set is a subset of the other).
What I currently have:
- Considering that $1 le n$, I "calculated" the arbitrary unions of $B_n$ to be:
$$bigcup_1 le n in mathbbNB_n = x in mathbbN ;$$ Proving the first direction:
Let:
$$ x in bigcup_1 le n in mathbbNB_n $$Since $1 le n in mathbbN$, then also $4 < x in mathbbN$.
Thus:
$$ bigcup_1 le n in mathbbNB_n subseteq x in mathbbN ; $$
Now, I'm not sure if my first-direction proof is correctly written - and please correct me if it's not - but what's more of an issue here is this:
How do I approach the opposite direction proof?
I'd be glad for any guidance.
elementary-set-theory proof-writing
elementary-set-theory proof-writing
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 32 mins ago
HeyJude
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 32 mins ago
HeyJude
1333
asked 32 mins ago
HeyJude
1333
1333
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
HeyJude is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $bigcup_1 le n in mathbbN B_n$ is equal to $ x in mathbbN mid 4 < x $, so let's look at the proof.
Your proof of the $subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$bigcup_i in I X_i = x mid x in X_i text for some i in I $$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $subseteq$ inclusion would look something like this:
Let $y in bigcup_1 le n in mathbbN B_n$. Then $y in B_n$ for some $n in mathbbN$ with $n ge 1$.
By definition of $B_n$, we have $3n+1 < y le 3n+4$.
Since $n ge 1$, we have $y > 4$, and so $y in x in mathbbN mid 4 < x $.
(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $supseteq$ direction, you need to let $y in x in mathbbN mid 4 < x $ and derive $y in bigcup_1 le n in mathbbN B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:
Let $y in x in mathbbN mid 4 < x $.
[...here you need to find (with proof) a value of $n in mathbbN$ with $n ge 1$ such that $y in B_n$...]
Hence $y in bigcup_1 le n in mathbbN B_n$.
The value of $n$ that you find will be given in terms of $y$.
answered 23 mins ago
Clive Newstead
48.9k471132
add a comment |Â
up vote
1
down vote
The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $xin B_n$ for some $nge1$, thus $4le 3n+1<x$.
For the other direction, given an $xinBbb N$ greater than $4$, how can you associate the $n$ to it, such that $xin B_n$? (Try small examples.)
answered 21 mins ago
Berci
58.5k23671
add a comment |Â
up vote
1
down vote
By evaluating the frist $B_n$'s,we get: $B_1=5,6,7$, $B_2=8,9,10$, $B_3=11,12,13$, so we see a pattern here. Each set has three consecutive numbers and, for each $ninmathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_n+1$ is $3(n+1)+2=3n+5$, so the first element of $B_n+1$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $ngeq5$.
answered 18 mins ago
André Porto
535213
add a comment |Â
up vote
0
down vote
Actually,
$$
bigcup_n=1^infty B_n=ninmathbb N: n>4.
$$
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $bigcup_1 le n in mathbbN B_n$ is equal to $ x in mathbbN mid 4 < x $, so let's look at the proof.
Your proof of the $subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$bigcup_i in I X_i = x mid x in X_i text for some i in I $$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $subseteq$ inclusion would look something like this:
Let $y in bigcup_1 le n in mathbbN B_n$. Then $y in B_n$ for some $n in mathbbN$ with $n ge 1$.
By definition of $B_n$, we have $3n+1 < y le 3n+4$.
Since $n ge 1$, we have $y > 4$, and so $y in x in mathbbN mid 4 < x $.
(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $supseteq$ direction, you need to let $y in x in mathbbN mid 4 < x $ and derive $y in bigcup_1 le n in mathbbN B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:
Let $y in x in mathbbN mid 4 < x $.
[...here you need to find (with proof) a value of $n in mathbbN$ with $n ge 1$ such that $y in B_n$...]
Hence $y in bigcup_1 le n in mathbbN B_n$.
The value of $n$ that you find will be given in terms of $y$.
answered 23 mins ago
Clive Newstead
48.9k471132
add a comment |Â
up vote
2
down vote
The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $bigcup_1 le n in mathbbN B_n$ is equal to $ x in mathbbN mid 4 < x $, so let's look at the proof.
Your proof of the $subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$bigcup_i in I X_i = x mid x in X_i text for some i in I $$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $subseteq$ inclusion would look something like this:
Let $y in bigcup_1 le n in mathbbN B_n$. Then $y in B_n$ for some $n in mathbbN$ with $n ge 1$.
By definition of $B_n$, we have $3n+1 < y le 3n+4$.
Since $n ge 1$, we have $y > 4$, and so $y in x in mathbbN mid 4 < x $.
(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $supseteq$ direction, you need to let $y in x in mathbbN mid 4 < x $ and derive $y in bigcup_1 le n in mathbbN B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:
Let $y in x in mathbbN mid 4 < x $.
[...here you need to find (with proof) a value of $n in mathbbN$ with $n ge 1$ such that $y in B_n$...]
Hence $y in bigcup_1 le n in mathbbN B_n$.
The value of $n$ that you find will be given in terms of $y$.
answered 23 mins ago
Clive Newstead
48.9k471132
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $bigcup_1 le n in mathbbN B_n$ is equal to $ x in mathbbN mid 4 < x $, so let's look at the proof.
Your proof of the $subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$bigcup_i in I X_i = x mid x in X_i text for some i in I $$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $subseteq$ inclusion would look something like this:
Let $y in bigcup_1 le n in mathbbN B_n$. Then $y in B_n$ for some $n in mathbbN$ with $n ge 1$.
By definition of $B_n$, we have $3n+1 < y le 3n+4$.
Since $n ge 1$, we have $y > 4$, and so $y in x in mathbbN mid 4 < x $.
(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $supseteq$ direction, you need to let $y in x in mathbbN mid 4 < x $ and derive $y in bigcup_1 le n in mathbbN B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:
Let $y in x in mathbbN mid 4 < x $.
[...here you need to find (with proof) a value of $n in mathbbN$ with $n ge 1$ such that $y in B_n$...]
Hence $y in bigcup_1 le n in mathbbN B_n$.
The value of $n$ that you find will be given in terms of $y$.
answered 23 mins ago
Clive Newstead
48.9k471132
The word calculate is fine—you could also say compute or evaluate, and you're correct that the union $bigcup_1 le n in mathbbN B_n$ is equal to $ x in mathbbN mid 4 < x $, so let's look at the proof.
Your proof of the $subseteq$ inclusion is correct but it is lacking some details. The definition of an indexed union is
$$bigcup_i in I X_i = x mid x in X_i text for some i in I $$
and this is really the definition you should be working with directly in your proof.
A more complete proof of the $subseteq$ inclusion would look something like this:
Let $y in bigcup_1 le n in mathbbN B_n$. Then $y in B_n$ for some $n in mathbbN$ with $n ge 1$.
By definition of $B_n$, we have $3n+1 < y le 3n+4$.
Since $n ge 1$, we have $y > 4$, and so $y in x in mathbbN mid 4 < x $.
(Notice that I used the variable $y$ to avoid overloading the variable $x$.)
For the $supseteq$ direction, you need to let $y in x in mathbbN mid 4 < x $ and derive $y in bigcup_1 le n in mathbbN B_n$, using the definition of the sets and set operations involved. Thus your proof should look like this:
Let $y in x in mathbbN mid 4 < x $.
[...here you need to find (with proof) a value of $n in mathbbN$ with $n ge 1$ such that $y in B_n$...]
Hence $y in bigcup_1 le n in mathbbN B_n$.
The value of $n$ that you find will be given in terms of $y$.
answered 23 mins ago
Clive Newstead
48.9k471132
answered 23 mins ago
Clive Newstead
48.9k471132
answered 23 mins ago
Clive Newstead
48.9k471132
answered 23 mins ago
Clive Newstead
48.9k471132
48.9k471132
add a comment |Â
add a comment |Â
up vote
1
down vote
The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $xin B_n$ for some $nge1$, thus $4le 3n+1<x$.
For the other direction, given an $xinBbb N$ greater than $4$, how can you associate the $n$ to it, such that $xin B_n$? (Try small examples.)
answered 21 mins ago
Berci
58.5k23671
add a comment |Â
up vote
1
down vote
The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $xin B_n$ for some $nge1$, thus $4le 3n+1<x$.
For the other direction, given an $xinBbb N$ greater than $4$, how can you associate the $n$ to it, such that $xin B_n$? (Try small examples.)
answered 21 mins ago
Berci
58.5k23671
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $xin B_n$ for some $nge1$, thus $4le 3n+1<x$.
For the other direction, given an $xinBbb N$ greater than $4$, how can you associate the $n$ to it, such that $xin B_n$? (Try small examples.)
answered 21 mins ago
Berci
58.5k23671
The word 'calculate' is ok, but you could also use e.g. 'find'.
The inclusion is correct, maybe you could insert a middle step: if $x$ is in the union of $B_n$'s, it means $xin B_n$ for some $nge1$, thus $4le 3n+1<x$.
For the other direction, given an $xinBbb N$ greater than $4$, how can you associate the $n$ to it, such that $xin B_n$? (Try small examples.)
answered 21 mins ago
Berci
58.5k23671
answered 21 mins ago
Berci
58.5k23671
answered 21 mins ago
Berci
58.5k23671
answered 21 mins ago
Berci
58.5k23671
58.5k23671
add a comment |Â
add a comment |Â
up vote
1
down vote
By evaluating the frist $B_n$'s,we get: $B_1=5,6,7$, $B_2=8,9,10$, $B_3=11,12,13$, so we see a pattern here. Each set has three consecutive numbers and, for each $ninmathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_n+1$ is $3(n+1)+2=3n+5$, so the first element of $B_n+1$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $ngeq5$.
answered 18 mins ago
André Porto
535213
add a comment |Â
up vote
1
down vote
By evaluating the frist $B_n$'s,we get: $B_1=5,6,7$, $B_2=8,9,10$, $B_3=11,12,13$, so we see a pattern here. Each set has three consecutive numbers and, for each $ninmathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_n+1$ is $3(n+1)+2=3n+5$, so the first element of $B_n+1$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $ngeq5$.
answered 18 mins ago
André Porto
535213
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By evaluating the frist $B_n$'s,we get: $B_1=5,6,7$, $B_2=8,9,10$, $B_3=11,12,13$, so we see a pattern here. Each set has three consecutive numbers and, for each $ninmathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_n+1$ is $3(n+1)+2=3n+5$, so the first element of $B_n+1$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $ngeq5$.
answered 18 mins ago
André Porto
535213
By evaluating the frist $B_n$'s,we get: $B_1=5,6,7$, $B_2=8,9,10$, $B_3=11,12,13$, so we see a pattern here. Each set has three consecutive numbers and, for each $ninmathbb N$, the last element of $B_n$ is $3n+4$ and the first element of $B_n+1$ is $3(n+1)+2=3n+5$, so the first element of $B_n+1$ is consecutive to the last element of $B_n$. Therefore, their union covers all natural numbers $ngeq5$.
answered 18 mins ago
André Porto
535213
answered 18 mins ago
André Porto
535213
answered 18 mins ago
André Porto
535213
answered 18 mins ago
André Porto
535213
535213
add a comment |Â
add a comment |Â
up vote
0
down vote
Actually,
$$
bigcup_n=1^infty B_n=ninmathbb N: n>4.
$$
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
add a comment |Â
up vote
0
down vote
Actually,
$$
bigcup_n=1^infty B_n=ninmathbb N: n>4.
$$
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Actually,
$$
bigcup_n=1^infty B_n=ninmathbb N: n>4.
$$
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
Actually,
$$
bigcup_n=1^infty B_n=ninmathbb N: n>4.
$$
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
answered 24 mins ago
Yiorgos S. Smyrlis
60.9k1383161
60.9k1383161
add a comment |Â
add a comment |Â
HeyJude is a new contributor. Be nice, and check out our Code of Conduct.
HeyJude is a new contributor. Be nice, and check out our Code of Conduct.
HeyJude is a new contributor. Be nice, and check out our Code of Conduct.
HeyJude is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2984232%2fprove-a-double-inclusion-of-an-arbitrary-unions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
