Prove $O = bigcup_{j=1}^infty O_j$ and $E subset bigcup_{j=1}^infty E_j implies O-E subset...
Prove that
$$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$
Below is my attempted proof, I'm stuck at the last expression.
Proof
$$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$
I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.
elementary-set-theory
asked Nov 26 '18 at 1:55
Zduff
1,544819
add a comment |
Prove that
$$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$
Below is my attempted proof, I'm stuck at the last expression.
Proof
$$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$
I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.
elementary-set-theory
asked Nov 26 '18 at 1:55
Zduff
1,544819
add a comment |
Prove that
$$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$
Below is my attempted proof, I'm stuck at the last expression.
Proof
$$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$
I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.
elementary-set-theory
asked Nov 26 '18 at 1:55
Zduff
1,544819
Prove that
$$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$
Below is my attempted proof, I'm stuck at the last expression.
Proof
$$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$
I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.
elementary-set-theory
elementary-set-theory
asked Nov 26 '18 at 1:55
Zduff
1,544819
asked Nov 26 '18 at 1:55
Zduff
1,544819
edited Nov 27 '18 at 1:52
asked Nov 26 '18 at 1:55
Zduff
1,544819
asked Nov 26 '18 at 1:55
Zduff
1,544819
asked Nov 26 '18 at 1:55
Zduff
1,544819
1,544819
add a comment |
add a comment |
1 Answer
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If $x in O - E$, then there exists $j$ such that $x in O_j$.
Then $x in O_j - E_j$ as well since $x notin E_j$.
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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active
oldest
votes
If $x in O - E$, then there exists $j$ such that $x in O_j$.
Then $x in O_j - E_j$ as well since $x notin E_j$.
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
add a comment |
If $x in O - E$, then there exists $j$ such that $x in O_j$.
Then $x in O_j - E_j$ as well since $x notin E_j$.
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
add a comment |
If $x in O - E$, then there exists $j$ such that $x in O_j$.
Then $x in O_j - E_j$ as well since $x notin E_j$.
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
If $x in O - E$, then there exists $j$ such that $x in O_j$.
Then $x in O_j - E_j$ as well since $x notin E_j$.
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
answered Nov 26 '18 at 1:57
angryavian
39.2k23280
39.2k23280
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
add a comment |
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
– DanielWainfleet
Nov 26 '18 at 7:01
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
@DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
– angryavian
Nov 26 '18 at 17:09
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
I expected it was a typo but I wanted to ask first.
– DanielWainfleet
Nov 28 '18 at 10:36
add a comment |
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