In which cases, there is no continuous map from A onto B?
4
$begingroup$
(a) $A=[0,1]cup[2,3], B=1,2$
(b) $A=(0,1), B=[0,1]$
(c) $A=mathbbQ, B=mathbbQ$
(d) $A=(0,1)cup(2,3), B=1,3$
It was clear for (b) as it was already asked numerous times on this site.
For (c), I took identity map.
For (d), We can send $(0,1)$ to $1$ and $(2,3)$ to $3$. Map is clearly onto and into a discrete space. It is continuous as inverse image of each singelton is open.
What about (a)?
real-analysis general-topology
edited Aug 28 '18 at 20:24
wchargin
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
$endgroup$
add a comment |
4
$begingroup$
(a) $A=[0,1]cup[2,3], B=1,2$
(b) $A=(0,1), B=[0,1]$
(c) $A=mathbbQ, B=mathbbQ$
(d) $A=(0,1)cup(2,3), B=1,3$
It was clear for (b) as it was already asked numerous times on this site.
For (c), I took identity map.
For (d), We can send $(0,1)$ to $1$ and $(2,3)$ to $3$. Map is clearly onto and into a discrete space. It is continuous as inverse image of each singelton is open.
What about (a)?
real-analysis general-topology
edited Aug 28 '18 at 20:24
wchargin
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
$endgroup$
1
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53
add a comment |
4
4
4
1
$begingroup$
(a) $A=[0,1]cup[2,3], B=1,2$
(b) $A=(0,1), B=[0,1]$
(c) $A=mathbbQ, B=mathbbQ$
(d) $A=(0,1)cup(2,3), B=1,3$
It was clear for (b) as it was already asked numerous times on this site.
For (c), I took identity map.
For (d), We can send $(0,1)$ to $1$ and $(2,3)$ to $3$. Map is clearly onto and into a discrete space. It is continuous as inverse image of each singelton is open.
What about (a)?
real-analysis general-topology
edited Aug 28 '18 at 20:24
wchargin
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
$endgroup$
(a) $A=[0,1]cup[2,3], B=1,2$
(b) $A=(0,1), B=[0,1]$
(c) $A=mathbbQ, B=mathbbQ$
(d) $A=(0,1)cup(2,3), B=1,3$
It was clear for (b) as it was already asked numerous times on this site.
For (c), I took identity map.
For (d), We can send $(0,1)$ to $1$ and $(2,3)$ to $3$. Map is clearly onto and into a discrete space. It is continuous as inverse image of each singelton is open.
What about (a)?
real-analysis general-topology
real-analysis general-topology
edited Aug 28 '18 at 20:24
wchargin
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
edited Aug 28 '18 at 20:24
wchargin
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
edited Aug 28 '18 at 20:24
wchargin
1,1161025
edited Aug 28 '18 at 20:24
wchargin
1,1161025
edited Aug 28 '18 at 20:24
wchargin
1,1161025
1,1161025
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
asked Aug 28 '18 at 15:25
StammeringMathematicianStammeringMathematician
2,6941324
2,6941324
1
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53
add a comment |
1
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53
1
1
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53
add a comment |
2 Answers
2
active
oldest
votes
7
$begingroup$
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
add a comment |
5
$begingroup$
a) goes the same as (d): map each interval to a separate point.
Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
add a comment |
7
$begingroup$
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
add a comment |
7
7
7
$begingroup$
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
answered Aug 28 '18 at 15:30
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
add a comment |
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
5
5
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
nice formula for b).
$endgroup$
– Henno Brandsma
Aug 28 '18 at 15:31
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
$begingroup$
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
$endgroup$
– StammeringMathematician
Aug 28 '18 at 16:18
2
2
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
$begingroup$
@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:21
2
2
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
$begingroup$
So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
$endgroup$
– Tanner Swett
Aug 28 '18 at 17:23
add a comment |
5
$begingroup$
a) goes the same as (d): map each interval to a separate point.
Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
add a comment |
5
$begingroup$
a) goes the same as (d): map each interval to a separate point.
Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
add a comment |
5
5
5
$begingroup$
a) goes the same as (d): map each interval to a separate point.
Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
a) goes the same as (d): map each interval to a separate point.
Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
answered Aug 28 '18 at 15:29
Henno BrandsmaHenno Brandsma
114k348123
114k348123
add a comment |
add a comment |
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1
$begingroup$
@phuclv changing letters to numbers invalidated existing answers.
$endgroup$
– Yakk
Aug 28 '18 at 17:51
$begingroup$
For (1) use the same idea as for (4)
$endgroup$
– DanielWainfleet
Aug 28 '18 at 17:52
$begingroup$
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
$endgroup$
– Yakk
Aug 28 '18 at 17:53