How to show Sup over bigger set is bigger and Inf over bigger set is lesser?
Let $A,B in mathbb{R}^n$ be two subset such that $A subseteq B$. Also, let $f:mathbb{R}^n rightarrow mathbb{R}$ be a real-valued function.
I always use
$$sup_Af(x) leq sup_Bf(x)$$
That perfectly makes sense to me, but is there any proof for the above inequality?
Also for the following
$$inf_Bf(x) leq inf_Af(x)$$
real-analysis functions supremum-and-infimum
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Let $A,B in mathbb{R}^n$ be two subset such that $A subseteq B$. Also, let $f:mathbb{R}^n rightarrow mathbb{R}$ be a real-valued function.
I always use
$$sup_Af(x) leq sup_Bf(x)$$
That perfectly makes sense to me, but is there any proof for the above inequality?
Also for the following
$$inf_Bf(x) leq inf_Af(x)$$
real-analysis functions supremum-and-infimum
add a comment |
Let $A,B in mathbb{R}^n$ be two subset such that $A subseteq B$. Also, let $f:mathbb{R}^n rightarrow mathbb{R}$ be a real-valued function.
I always use
$$sup_Af(x) leq sup_Bf(x)$$
That perfectly makes sense to me, but is there any proof for the above inequality?
Also for the following
$$inf_Bf(x) leq inf_Af(x)$$
real-analysis functions supremum-and-infimum
Let $A,B in mathbb{R}^n$ be two subset such that $A subseteq B$. Also, let $f:mathbb{R}^n rightarrow mathbb{R}$ be a real-valued function.
I always use
$$sup_Af(x) leq sup_Bf(x)$$
That perfectly makes sense to me, but is there any proof for the above inequality?
Also for the following
$$inf_Bf(x) leq inf_Af(x)$$
real-analysis functions supremum-and-infimum
real-analysis functions supremum-and-infimum
asked Nov 25 at 19:08
Saeed
585110
585110
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2 Answers
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Hint: $sup_{xin B}f(x)$ is an upper bound of the set ${f(x),|,xin A}$.
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Let $xin A$. Then in particular $xin B$ and $f(x)geq inf_{xin B}f(x)$. So $inf_{xin B}f(x)$ is a lower bound for the values of $f(x)$ on $xin A$. In particular $inf_{xin A}f(x)geq inf_{xin B}f(x)$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: $sup_{xin B}f(x)$ is an upper bound of the set ${f(x),|,xin A}$.
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Hint: $sup_{xin B}f(x)$ is an upper bound of the set ${f(x),|,xin A}$.
add a comment |
Hint: $sup_{xin B}f(x)$ is an upper bound of the set ${f(x),|,xin A}$.
Hint: $sup_{xin B}f(x)$ is an upper bound of the set ${f(x),|,xin A}$.
answered Nov 25 at 19:11
José Carlos Santos
148k22117218
148k22117218
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Let $xin A$. Then in particular $xin B$ and $f(x)geq inf_{xin B}f(x)$. So $inf_{xin B}f(x)$ is a lower bound for the values of $f(x)$ on $xin A$. In particular $inf_{xin A}f(x)geq inf_{xin B}f(x)$.
add a comment |
Let $xin A$. Then in particular $xin B$ and $f(x)geq inf_{xin B}f(x)$. So $inf_{xin B}f(x)$ is a lower bound for the values of $f(x)$ on $xin A$. In particular $inf_{xin A}f(x)geq inf_{xin B}f(x)$.
add a comment |
Let $xin A$. Then in particular $xin B$ and $f(x)geq inf_{xin B}f(x)$. So $inf_{xin B}f(x)$ is a lower bound for the values of $f(x)$ on $xin A$. In particular $inf_{xin A}f(x)geq inf_{xin B}f(x)$.
Let $xin A$. Then in particular $xin B$ and $f(x)geq inf_{xin B}f(x)$. So $inf_{xin B}f(x)$ is a lower bound for the values of $f(x)$ on $xin A$. In particular $inf_{xin A}f(x)geq inf_{xin B}f(x)$.
answered Nov 25 at 19:11
Foobaz John
20.8k41250
20.8k41250
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