Suppose $x$ is a non-negative real number such that for all $epsilon >0$ we have $x < epsilon$. Then...
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Is the following statement true?
Suppose $x$ is a nonnegative real number such that for all $epsilon >0$ we have $x < epsilon$. Then $x=0$.
Intuitively speaking yes since $0$ is an infimum of the set ${epsilon space|epsilon >0}$. I was trying to come up with an axiomatic reason for this but all I could find was algebraic properties of $0$ on $mathbb{R}$ which didn't seem too relevant. But can this result be derived from the order axioms of $mathbb{R}$ if true?
real-numbers ordered-fields
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
add a comment |
up vote
1
down vote
favorite
Is the following statement true?
Suppose $x$ is a nonnegative real number such that for all $epsilon >0$ we have $x < epsilon$. Then $x=0$.
Intuitively speaking yes since $0$ is an infimum of the set ${epsilon space|epsilon >0}$. I was trying to come up with an axiomatic reason for this but all I could find was algebraic properties of $0$ on $mathbb{R}$ which didn't seem too relevant. But can this result be derived from the order axioms of $mathbb{R}$ if true?
real-numbers ordered-fields
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is the following statement true?
Suppose $x$ is a nonnegative real number such that for all $epsilon >0$ we have $x < epsilon$. Then $x=0$.
Intuitively speaking yes since $0$ is an infimum of the set ${epsilon space|epsilon >0}$. I was trying to come up with an axiomatic reason for this but all I could find was algebraic properties of $0$ on $mathbb{R}$ which didn't seem too relevant. But can this result be derived from the order axioms of $mathbb{R}$ if true?
real-numbers ordered-fields
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
Is the following statement true?
Suppose $x$ is a nonnegative real number such that for all $epsilon >0$ we have $x < epsilon$. Then $x=0$.
Intuitively speaking yes since $0$ is an infimum of the set ${epsilon space|epsilon >0}$. I was trying to come up with an axiomatic reason for this but all I could find was algebraic properties of $0$ on $mathbb{R}$ which didn't seem too relevant. But can this result be derived from the order axioms of $mathbb{R}$ if true?
real-numbers ordered-fields
real-numbers ordered-fields
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
edited Nov 14 at 21:55
José Carlos Santos
140k19111204
140k19111204
asked Nov 14 at 21:46
Sei Sakata
567
asked Nov 14 at 21:46
Sei Sakata
567
asked Nov 14 at 21:46
Sei Sakata
567
567
add a comment |
add a comment |
1 Answer
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If $x>0$, take $varepsilon=x$. Then $x<varepsilon=x$, which is impossible.
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $x>0$, take $varepsilon=x$. Then $x<varepsilon=x$, which is impossible.
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
add a comment |
up vote
2
down vote
If $x>0$, take $varepsilon=x$. Then $x<varepsilon=x$, which is impossible.
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
add a comment |
up vote
2
down vote
up vote
2
down vote
If $x>0$, take $varepsilon=x$. Then $x<varepsilon=x$, which is impossible.
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
If $x>0$, take $varepsilon=x$. Then $x<varepsilon=x$, which is impossible.
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
answered Nov 14 at 21:48
José Carlos Santos
140k19111204
140k19111204
add a comment |
add a comment |
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