How to show a continuous function in the reals maps compact sets into compact sets?
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How would I go about proving a continuous function maps compact sets into compact sets for the reals? I have seen this question pointing me to the Intermediate Value Theorem wiki, but I am still confused. Other answers use topology terms and inverse images, but I don't have those defined yet in my course.
real-analysis compactness
asked Dec 18 '18 at 13:25
kaisakaisa
2019
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add a comment |
$begingroup$
How would I go about proving a continuous function maps compact sets into compact sets for the reals? I have seen this question pointing me to the Intermediate Value Theorem wiki, but I am still confused. Other answers use topology terms and inverse images, but I don't have those defined yet in my course.
real-analysis compactness
asked Dec 18 '18 at 13:25
kaisakaisa
2019
$endgroup$
$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33
add a comment |
$begingroup$
How would I go about proving a continuous function maps compact sets into compact sets for the reals? I have seen this question pointing me to the Intermediate Value Theorem wiki, but I am still confused. Other answers use topology terms and inverse images, but I don't have those defined yet in my course.
real-analysis compactness
asked Dec 18 '18 at 13:25
kaisakaisa
2019
$endgroup$
How would I go about proving a continuous function maps compact sets into compact sets for the reals? I have seen this question pointing me to the Intermediate Value Theorem wiki, but I am still confused. Other answers use topology terms and inverse images, but I don't have those defined yet in my course.
real-analysis compactness
real-analysis compactness
asked Dec 18 '18 at 13:25
kaisakaisa
2019
asked Dec 18 '18 at 13:25
kaisakaisa
2019
asked Dec 18 '18 at 13:25
kaisakaisa
2019
asked Dec 18 '18 at 13:25
kaisakaisa
2019
asked Dec 18 '18 at 13:25
kaisakaisa
2019
2019
$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33
add a comment |
$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33
$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33
$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33
add a comment |
1 Answer
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Hint
Start with an open cover of the image, then using the fact that the function is continuous shows the preimage of that cover is open and covers the domain (i.e the compact set), therefore it has an open subcover. It should follow easily from there
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Start with an open cover of the image, then using the fact that the function is continuous shows the preimage of that cover is open and covers the domain (i.e the compact set), therefore it has an open subcover. It should follow easily from there
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
$endgroup$
add a comment |
$begingroup$
Hint
Start with an open cover of the image, then using the fact that the function is continuous shows the preimage of that cover is open and covers the domain (i.e the compact set), therefore it has an open subcover. It should follow easily from there
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
$endgroup$
add a comment |
$begingroup$
Hint
Start with an open cover of the image, then using the fact that the function is continuous shows the preimage of that cover is open and covers the domain (i.e the compact set), therefore it has an open subcover. It should follow easily from there
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
$endgroup$
Hint
Start with an open cover of the image, then using the fact that the function is continuous shows the preimage of that cover is open and covers the domain (i.e the compact set), therefore it has an open subcover. It should follow easily from there
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
answered Dec 18 '18 at 13:39
Scosh_lrScosh_lr
516
516
add a comment |
add a comment |
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$begingroup$
What's the concept of “compact” that you are working with?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 13:33