union of finite bounded set and uniformly bounded set is bounded
$begingroup$
Let $A, G subset C ([a, b])$, $G = {g_1, g_2, ..., g_m}$ (finite set).
Prove that if: i) $A || .. ||$ $infty$-bounded then $A cup G$ too.
ii) $A$ equicontinuous in $x_o$ then $A cup G$ also.
for i) since $G$ is finite it has a max and a min element. I tried the triangle inequality taking the distance of any two element of $A$ and $G$.
ii) Don't know how to proceed.
general-topology arzela-ascoli
edited Dec 8 '18 at 5:02
twnly
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
$endgroup$
add a comment |
$begingroup$
Let $A, G subset C ([a, b])$, $G = {g_1, g_2, ..., g_m}$ (finite set).
Prove that if: i) $A || .. ||$ $infty$-bounded then $A cup G$ too.
ii) $A$ equicontinuous in $x_o$ then $A cup G$ also.
for i) since $G$ is finite it has a max and a min element. I tried the triangle inequality taking the distance of any two element of $A$ and $G$.
ii) Don't know how to proceed.
general-topology arzela-ascoli
edited Dec 8 '18 at 5:02
twnly
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
$endgroup$
add a comment |
$begingroup$
Let $A, G subset C ([a, b])$, $G = {g_1, g_2, ..., g_m}$ (finite set).
Prove that if: i) $A || .. ||$ $infty$-bounded then $A cup G$ too.
ii) $A$ equicontinuous in $x_o$ then $A cup G$ also.
for i) since $G$ is finite it has a max and a min element. I tried the triangle inequality taking the distance of any two element of $A$ and $G$.
ii) Don't know how to proceed.
general-topology arzela-ascoli
edited Dec 8 '18 at 5:02
twnly
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
$endgroup$
Let $A, G subset C ([a, b])$, $G = {g_1, g_2, ..., g_m}$ (finite set).
Prove that if: i) $A || .. ||$ $infty$-bounded then $A cup G$ too.
ii) $A$ equicontinuous in $x_o$ then $A cup G$ also.
for i) since $G$ is finite it has a max and a min element. I tried the triangle inequality taking the distance of any two element of $A$ and $G$.
ii) Don't know how to proceed.
general-topology arzela-ascoli
general-topology arzela-ascoli
edited Dec 8 '18 at 5:02
twnly
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
edited Dec 8 '18 at 5:02
twnly
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
edited Dec 8 '18 at 5:02
twnly
1,1221213
edited Dec 8 '18 at 5:02
twnly
1,1221213
edited Dec 8 '18 at 5:02
twnly
1,1221213
1,1221213
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
asked Dec 8 '18 at 4:11
sofia de la morasofia de la mora
31
31
add a comment |
add a comment |
1 Answer
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$begingroup$
Show from the definitions:
A union of two $infty$-bounded sets is $infty$-bounded.
A finite set is $infty$-bounded.
Show the same two facts for equicontinuous sets.
For the finite case use that a single continuous function on $[a,b]$ is bounded and uniformly continuous.
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Show from the definitions:
A union of two $infty$-bounded sets is $infty$-bounded.
A finite set is $infty$-bounded.
Show the same two facts for equicontinuous sets.
For the finite case use that a single continuous function on $[a,b]$ is bounded and uniformly continuous.
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
Show from the definitions:
A union of two $infty$-bounded sets is $infty$-bounded.
A finite set is $infty$-bounded.
Show the same two facts for equicontinuous sets.
For the finite case use that a single continuous function on $[a,b]$ is bounded and uniformly continuous.
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
Show from the definitions:
A union of two $infty$-bounded sets is $infty$-bounded.
A finite set is $infty$-bounded.
Show the same two facts for equicontinuous sets.
For the finite case use that a single continuous function on $[a,b]$ is bounded and uniformly continuous.
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
Show from the definitions:
A union of two $infty$-bounded sets is $infty$-bounded.
A finite set is $infty$-bounded.
Show the same two facts for equicontinuous sets.
For the finite case use that a single continuous function on $[a,b]$ is bounded and uniformly continuous.
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
answered Dec 8 '18 at 5:19
Henno BrandsmaHenno Brandsma
112k348121
112k348121
add a comment |
add a comment |
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