In which cases, there is no continuous map from A onto B?

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(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)?







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  • 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















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)?







share|cite|improve this question


















  • 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













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)?







share|cite|improve this question














(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)?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 28 at 20:24









wchargin

1,0981024




1,0981024










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













  • 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











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) $xmapsto lceil xrceil$






share|cite|improve this answer
















  • 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

















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 $[0,1] cup [2,3]$.






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    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) $xmapsto lceil xrceil$






    share|cite|improve this answer
















    • 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














    up vote
    7
    down vote



    accepted










    a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$



    b) $xmapsto frac1+sin 42x2$



    c) $xmapsto x$



    d) $xmapsto lceil xrceil$






    share|cite|improve this answer
















    • 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












    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) $xmapsto lceil xrceil$






    share|cite|improve this answer












    a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$



    b) $xmapsto frac1+sin 42x2$



    c) $xmapsto x$



    d) $xmapsto lceil xrceil$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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












    • 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










    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 $[0,1] cup [2,3]$.






    share|cite|improve this answer
























      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 $[0,1] cup [2,3]$.






      share|cite|improve this answer






















        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 $[0,1] cup [2,3]$.






        share|cite|improve this answer












        a) goes the same as (d): map each interval to a separate point.



        Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 28 at 15:29









        Henno Brandsma

        92.9k342100




        92.9k342100



























             

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