Proof verification for bounded sets.
$begingroup$
Is my proof for set boundedness correct?
Proposition:
Let $Asubseteq mathbb{R}.$ Then $A$ is bounded if and only if $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$.
Proof:
Suppose that $A$ is bounded. Then there exists $Min mathbb{R}$ and $minmathbb{R}$ such that $mleq{a}leq{M}$. Take $K=text{max}(|M|,|m|,1)$ so that $Kgt{0}$. Therefore $-Kle{a}le{K}$ and thus, $|a|le{K}$.
Conversely, assume that $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$ holds. Then we have that $-Kleq{a}leq{K}$. Therefore $-Kleq{a}$ for all $ain{A}$ so $A$ is bounded below and $aleq{K}$
for all $ain{A}$ so $A$ is bounded above. Thus, $A$ is bounded.
proof-verification
$endgroup$
add a comment |
$begingroup$
Is my proof for set boundedness correct?
Proposition:
Let $Asubseteq mathbb{R}.$ Then $A$ is bounded if and only if $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$.
Proof:
Suppose that $A$ is bounded. Then there exists $Min mathbb{R}$ and $minmathbb{R}$ such that $mleq{a}leq{M}$. Take $K=text{max}(|M|,|m|,1)$ so that $Kgt{0}$. Therefore $-Kle{a}le{K}$ and thus, $|a|le{K}$.
Conversely, assume that $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$ holds. Then we have that $-Kleq{a}leq{K}$. Therefore $-Kleq{a}$ for all $ain{A}$ so $A$ is bounded below and $aleq{K}$
for all $ain{A}$ so $A$ is bounded above. Thus, $A$ is bounded.
proof-verification
$endgroup$
4
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
2
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
2
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17
add a comment |
$begingroup$
Is my proof for set boundedness correct?
Proposition:
Let $Asubseteq mathbb{R}.$ Then $A$ is bounded if and only if $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$.
Proof:
Suppose that $A$ is bounded. Then there exists $Min mathbb{R}$ and $minmathbb{R}$ such that $mleq{a}leq{M}$. Take $K=text{max}(|M|,|m|,1)$ so that $Kgt{0}$. Therefore $-Kle{a}le{K}$ and thus, $|a|le{K}$.
Conversely, assume that $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$ holds. Then we have that $-Kleq{a}leq{K}$. Therefore $-Kleq{a}$ for all $ain{A}$ so $A$ is bounded below and $aleq{K}$
for all $ain{A}$ so $A$ is bounded above. Thus, $A$ is bounded.
proof-verification
$endgroup$
Is my proof for set boundedness correct?
Proposition:
Let $Asubseteq mathbb{R}.$ Then $A$ is bounded if and only if $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$.
Proof:
Suppose that $A$ is bounded. Then there exists $Min mathbb{R}$ and $minmathbb{R}$ such that $mleq{a}leq{M}$. Take $K=text{max}(|M|,|m|,1)$ so that $Kgt{0}$. Therefore $-Kle{a}le{K}$ and thus, $|a|le{K}$.
Conversely, assume that $exists{K}gt{0}$ s.t $forall{a}in{A}, |a|leq{K}$ holds. Then we have that $-Kleq{a}leq{K}$. Therefore $-Kleq{a}$ for all $ain{A}$ so $A$ is bounded below and $aleq{K}$
for all $ain{A}$ so $A$ is bounded above. Thus, $A$ is bounded.
proof-verification
proof-verification
asked Jan 22 at 14:39
user503154
4
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
2
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
2
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17
add a comment |
4
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
2
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
2
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17
4
4
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
2
2
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
2
2
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17
add a comment |
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4
$begingroup$
Yes it is correct.
$endgroup$
– Yanko
Jan 22 at 14:40
2
$begingroup$
Approved. $ddotsmile$
$endgroup$
– Maksim
Jan 22 at 15:13
2
$begingroup$
Thanks to you both ;)
$endgroup$
– user503154
Jan 22 at 15:17