In which cases, there is no continuous map from A onto B?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
1
(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 at 20:24
wchargin
1,0981024
asked Aug 28 at 15:25
StammeringMathematician
32810
add a comment |Â
up vote
4
down vote
favorite
1
(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 at 20:24
wchargin
1,0981024
asked Aug 28 at 15:25
StammeringMathematician
32810
1
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53
add a comment |Â
up vote
4
down vote
favorite
1
up vote
4
down vote
favorite
1
1
(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 at 20:24
wchargin
1,0981024
asked Aug 28 at 15:25
StammeringMathematician
32810
(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 at 20:24
wchargin
1,0981024
asked Aug 28 at 15:25
StammeringMathematician
32810
edited Aug 28 at 20:24
wchargin
1,0981024
edited Aug 28 at 20:24
wchargin
1,0981024
edited Aug 28 at 20:24
wchargin
1,0981024
1,0981024
asked Aug 28 at 15:25
StammeringMathematician
32810
asked Aug 28 at 15:25
StammeringMathematician
32810
asked Aug 28 at 15:25
StammeringMathematician
32810
32810
1
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53
add a comment |Â
1
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53
1
1
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
@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...)
– Tanner Swett
Aug 28 at 17:21
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
add a comment |Â
up vote
5
down vote
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 at 15:29
Henno Brandsma
92.9k342100
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
@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...)
– Tanner Swett
Aug 28 at 17:21
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
add a comment |Â
up vote
7
down vote
accepted
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
@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...)
– Tanner Swett
Aug 28 at 17:21
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$
b) $xmapsto frac1+sin 42x2$
c) $xmapsto x$
d) $xmapsto lceil xrceil$
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
answered Aug 28 at 15:30
Hagen von Eitzen
267k21259481
267k21259481
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
@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...)
– Tanner Swett
Aug 28 at 17:21
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
add a comment |Â
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
@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...)
– Tanner Swett
Aug 28 at 17:21
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
5
5
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
nice formula for b).
– Henno Brandsma
Aug 28 at 15:31
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18
1
1
@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...)
– Tanner Swett
Aug 28 at 17:21
@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...)
– Tanner Swett
Aug 28 at 17:21
1
1
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
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 $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23
add a comment |Â
up vote
5
down vote
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 at 15:29
Henno Brandsma
92.9k342100
add a comment |Â
up vote
5
down vote
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 at 15:29
Henno Brandsma
92.9k342100
add a comment |Â
up vote
5
down vote
up vote
5
down vote
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 at 15:29
Henno Brandsma
92.9k342100
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 at 15:29
Henno Brandsma
92.9k342100
answered Aug 28 at 15:29
Henno Brandsma
92.9k342100
answered Aug 28 at 15:29
Henno Brandsma
92.9k342100
answered Aug 28 at 15:29
Henno Brandsma
92.9k342100
92.9k342100
add a comment |Â
add a comment |Â
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2897396%2fin-which-cases-there-is-no-continuous-map-from-a-onto-b%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51
For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52
I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53