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A sequence of nested unbounded closed intervals $L_1supseteq L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n = varnothing$
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Is the following argument correct?
Prove that there exists a sequence of nested unbounded closed intervals $L_1supseteq
L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n =
varnothing$. (An unbounded closed interval has the form $[a,infty) = xinmathbfR:xge a$).
Solution. Consider the nested intervals $L_1 = [1,infty),L_2 = [2,infty),L_3 = [3,infty)cdots$, Now assume that we have an $xinmathbfR$ such that $xin bigcap_n=1^inftyL_n$ and let $kinmathbfN$, from hypothesis $xin L_k = [k,infty)$ then $xge k$ but $xneq k$ as that would imply that $xnotin L_k+1$, therefore it must be that $x>k$, then since $k$ was arbitrary it follows that $x$ is an upper bound for $mathbfN$, but this is impossible since we know that the real numbers possess the archimedian property, consequently $bigcap_n=1^inftyL_n = varnothing$.
$blacksquare$
real-analysis proof-verification
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
asked Jul 22 at 17:51
Atif Farooq
2,7092824
add a comment |Â
up vote
2
down vote
favorite
Is the following argument correct?
Prove that there exists a sequence of nested unbounded closed intervals $L_1supseteq
L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n =
varnothing$. (An unbounded closed interval has the form $[a,infty) = xinmathbfR:xge a$).
Solution. Consider the nested intervals $L_1 = [1,infty),L_2 = [2,infty),L_3 = [3,infty)cdots$, Now assume that we have an $xinmathbfR$ such that $xin bigcap_n=1^inftyL_n$ and let $kinmathbfN$, from hypothesis $xin L_k = [k,infty)$ then $xge k$ but $xneq k$ as that would imply that $xnotin L_k+1$, therefore it must be that $x>k$, then since $k$ was arbitrary it follows that $x$ is an upper bound for $mathbfN$, but this is impossible since we know that the real numbers possess the archimedian property, consequently $bigcap_n=1^inftyL_n = varnothing$.
$blacksquare$
real-analysis proof-verification
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
asked Jul 22 at 17:51
Atif Farooq
2,7092824
1
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is the following argument correct?
Prove that there exists a sequence of nested unbounded closed intervals $L_1supseteq
L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n =
varnothing$. (An unbounded closed interval has the form $[a,infty) = xinmathbfR:xge a$).
Solution. Consider the nested intervals $L_1 = [1,infty),L_2 = [2,infty),L_3 = [3,infty)cdots$, Now assume that we have an $xinmathbfR$ such that $xin bigcap_n=1^inftyL_n$ and let $kinmathbfN$, from hypothesis $xin L_k = [k,infty)$ then $xge k$ but $xneq k$ as that would imply that $xnotin L_k+1$, therefore it must be that $x>k$, then since $k$ was arbitrary it follows that $x$ is an upper bound for $mathbfN$, but this is impossible since we know that the real numbers possess the archimedian property, consequently $bigcap_n=1^inftyL_n = varnothing$.
$blacksquare$
real-analysis proof-verification
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
asked Jul 22 at 17:51
Atif Farooq
2,7092824
Is the following argument correct?
Prove that there exists a sequence of nested unbounded closed intervals $L_1supseteq
L_2supseteq L_3supseteqcdots$ with $bigcap_n=1^inftyL_n =
varnothing$. (An unbounded closed interval has the form $[a,infty) = xinmathbfR:xge a$).
Solution. Consider the nested intervals $L_1 = [1,infty),L_2 = [2,infty),L_3 = [3,infty)cdots$, Now assume that we have an $xinmathbfR$ such that $xin bigcap_n=1^inftyL_n$ and let $kinmathbfN$, from hypothesis $xin L_k = [k,infty)$ then $xge k$ but $xneq k$ as that would imply that $xnotin L_k+1$, therefore it must be that $x>k$, then since $k$ was arbitrary it follows that $x$ is an upper bound for $mathbfN$, but this is impossible since we know that the real numbers possess the archimedian property, consequently $bigcap_n=1^inftyL_n = varnothing$.
$blacksquare$
real-analysis proof-verification
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
asked Jul 22 at 17:51
Atif Farooq
2,7092824
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
edited Jul 22 at 19:48
DanielWainfleet
31.6k31543
31.6k31543
asked Jul 22 at 17:51
Atif Farooq
2,7092824
asked Jul 22 at 17:51
Atif Farooq
2,7092824
asked Jul 22 at 17:51
Atif Farooq
2,7092824
2,7092824
1
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52
add a comment |Â
1
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52
1
1
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52
add a comment |Â
1 Answer
1
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votes
up vote
2
down vote
accepted
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_k+1$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $mathbfN$, which is found by $xge k$
edited Jul 22 at 21:07
Atif Farooq
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_k+1$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $mathbfN$, which is found by $xge k$
edited Jul 22 at 21:07
Atif Farooq
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
accepted
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_k+1$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $mathbfN$, which is found by $xge k$
edited Jul 22 at 21:07
Atif Farooq
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_k+1$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $mathbfN$, which is found by $xge k$
edited Jul 22 at 21:07
Atif Farooq
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_k+1$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $mathbfN$, which is found by $xge k$
edited Jul 22 at 21:07
Atif Farooq
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
edited Jul 22 at 21:07
Atif Farooq
2,7092824
edited Jul 22 at 21:07
Atif Farooq
2,7092824
edited Jul 22 at 21:07
Atif Farooq
2,7092824
2,7092824
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 19:00
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
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1
Did you mean: "There exists a sequence of nested unbounded closed intervals..."?
– Sambo
Jul 22 at 17:55
@Sambo Yes i did, my apologies for the sloppiness
– Atif Farooq
Jul 22 at 17:55
Then yes, your argument is correct. Though I'm not sure why you needed to prove $x>k$; it seems to me that $x geq k$ would be sufficient.
– Sambo
Jul 22 at 17:57
Correct although a bit wordy. You could say that if $x$ belongs to every $L_k$ than $xgeq k$ for every $kin Bbb N$, which is absurd..... My edit was to change $J$ to $L$ throughout.
– DanielWainfleet
Jul 22 at 19:52