Prove property of $A$ implies $inf A le 2$
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Set $A$ is a proper subset of reals and has a property that for every $a in A$ there exists $b in A$ such that $a ge 2b - 2$. How to prove that $inf A le 2$?
supremum-and-infimum
asked Nov 17 at 13:12
user4201961
625411
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Set $A$ is a proper subset of reals and has a property that for every $a in A$ there exists $b in A$ such that $a ge 2b - 2$. How to prove that $inf A le 2$?
supremum-and-infimum
asked Nov 17 at 13:12
user4201961
625411
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Set $A$ is a proper subset of reals and has a property that for every $a in A$ there exists $b in A$ such that $a ge 2b - 2$. How to prove that $inf A le 2$?
supremum-and-infimum
asked Nov 17 at 13:12
user4201961
625411
Set $A$ is a proper subset of reals and has a property that for every $a in A$ there exists $b in A$ such that $a ge 2b - 2$. How to prove that $inf A le 2$?
supremum-and-infimum
supremum-and-infimum
asked Nov 17 at 13:12
user4201961
625411
asked Nov 17 at 13:12
user4201961
625411
edited Nov 17 at 13:17
asked Nov 17 at 13:12
user4201961
625411
asked Nov 17 at 13:12
user4201961
625411
asked Nov 17 at 13:12
user4201961
625411
625411
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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Hint:
If $2b-2leq a$ then $bleq frac{a+2}{2}$ (so $b$ is less than or equal the average between $a$ and $2$).
Moreover, if $a>2$ then $frac{a+2}{2}<a$.
We conclude that if $a>2$ then there exists $bin A$ such that $b<a$ can you finish now?
answered Nov 17 at 13:49
Yanko
5,018722
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:
If $2b-2leq a$ then $bleq frac{a+2}{2}$ (so $b$ is less than or equal the average between $a$ and $2$).
Moreover, if $a>2$ then $frac{a+2}{2}<a$.
We conclude that if $a>2$ then there exists $bin A$ such that $b<a$ can you finish now?
answered Nov 17 at 13:49
Yanko
5,018722
add a comment |
up vote
1
down vote
accepted
Hint:
If $2b-2leq a$ then $bleq frac{a+2}{2}$ (so $b$ is less than or equal the average between $a$ and $2$).
Moreover, if $a>2$ then $frac{a+2}{2}<a$.
We conclude that if $a>2$ then there exists $bin A$ such that $b<a$ can you finish now?
answered Nov 17 at 13:49
Yanko
5,018722
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:
If $2b-2leq a$ then $bleq frac{a+2}{2}$ (so $b$ is less than or equal the average between $a$ and $2$).
Moreover, if $a>2$ then $frac{a+2}{2}<a$.
We conclude that if $a>2$ then there exists $bin A$ such that $b<a$ can you finish now?
answered Nov 17 at 13:49
Yanko
5,018722
Hint:
If $2b-2leq a$ then $bleq frac{a+2}{2}$ (so $b$ is less than or equal the average between $a$ and $2$).
Moreover, if $a>2$ then $frac{a+2}{2}<a$.
We conclude that if $a>2$ then there exists $bin A$ such that $b<a$ can you finish now?
answered Nov 17 at 13:49
Yanko
5,018722
answered Nov 17 at 13:49
Yanko
5,018722
answered Nov 17 at 13:49
Yanko
5,018722
answered Nov 17 at 13:49
Yanko
5,018722
5,018722
add a comment |
add a comment |
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