Euclidean topology: preimage of $(a, b)$ with $a < b leq 0$












0














I am encountering some difficulties and would like to ask for a hint.
I have a continuous function$$f:Rrightarrow R: f(x)=x^2$$
and am supposed to find preimages of the following intervals



$$1.(a,b) , 0leq a< b $$
$$2.(a,b) , a < b leq 0$$
$$3.(a,b) , a leq 0 <b$$



I am wondering how the function could spit out a negative value as needed in 2. where a is supposed to be strictly smaller than 0.



I apologize, i should add that I do not believe equation 2 to have a preimage, however part 4 of the exercise is then the following:



With the previous results show that $f$ is (topologically) continuous. (Recall that any open set in R can be represented as a union of intervals of the form (a, b) with a < b.).



As I understand, $f(x)=x^2$ is continuous topologically and therefore i don't believe my answer of equation 2 not having a preimage to be correct. Would you be so kind to explain if my understanding is incorrect?






general-topology functions continuity







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  • 1




    It couldn't. Therefore...
    – user3482749
    Dec 4 '18 at 15:37
















0














I am encountering some difficulties and would like to ask for a hint.
I have a continuous function$$f:Rrightarrow R: f(x)=x^2$$
and am supposed to find preimages of the following intervals



$$1.(a,b) , 0leq a< b $$
$$2.(a,b) , a < b leq 0$$
$$3.(a,b) , a leq 0 <b$$



I am wondering how the function could spit out a negative value as needed in 2. where a is supposed to be strictly smaller than 0.



I apologize, i should add that I do not believe equation 2 to have a preimage, however part 4 of the exercise is then the following:



With the previous results show that $f$ is (topologically) continuous. (Recall that any open set in R can be represented as a union of intervals of the form (a, b) with a < b.).



As I understand, $f(x)=x^2$ is continuous topologically and therefore i don't believe my answer of equation 2 not having a preimage to be correct. Would you be so kind to explain if my understanding is incorrect?






general-topology functions continuity







share|cite|improve this question




















  • 1




    It couldn't. Therefore...
    – user3482749
    Dec 4 '18 at 15:37














0












0








0







I am encountering some difficulties and would like to ask for a hint.
I have a continuous function$$f:Rrightarrow R: f(x)=x^2$$
and am supposed to find preimages of the following intervals



$$1.(a,b) , 0leq a< b $$
$$2.(a,b) , a < b leq 0$$
$$3.(a,b) , a leq 0 <b$$



I am wondering how the function could spit out a negative value as needed in 2. where a is supposed to be strictly smaller than 0.



I apologize, i should add that I do not believe equation 2 to have a preimage, however part 4 of the exercise is then the following:



With the previous results show that $f$ is (topologically) continuous. (Recall that any open set in R can be represented as a union of intervals of the form (a, b) with a < b.).



As I understand, $f(x)=x^2$ is continuous topologically and therefore i don't believe my answer of equation 2 not having a preimage to be correct. Would you be so kind to explain if my understanding is incorrect?






general-topology functions continuity







share|cite|improve this question















I am encountering some difficulties and would like to ask for a hint.
I have a continuous function$$f:Rrightarrow R: f(x)=x^2$$
and am supposed to find preimages of the following intervals



$$1.(a,b) , 0leq a< b $$
$$2.(a,b) , a < b leq 0$$
$$3.(a,b) , a leq 0 <b$$



I am wondering how the function could spit out a negative value as needed in 2. where a is supposed to be strictly smaller than 0.



I apologize, i should add that I do not believe equation 2 to have a preimage, however part 4 of the exercise is then the following:



With the previous results show that $f$ is (topologically) continuous. (Recall that any open set in R can be represented as a union of intervals of the form (a, b) with a < b.).



As I understand, $f(x)=x^2$ is continuous topologically and therefore i don't believe my answer of equation 2 not having a preimage to be correct. Would you be so kind to explain if my understanding is incorrect?






general-topology functions continuity




general-topology functions continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 15:49

























asked Dec 4 '18 at 15:36









maruto

33




33








  • 1




    It couldn't. Therefore...
    – user3482749
    Dec 4 '18 at 15:37














  • 1




    It couldn't. Therefore...
    – user3482749
    Dec 4 '18 at 15:37








1




1




It couldn't. Therefore...
– user3482749
Dec 4 '18 at 15:37




It couldn't. Therefore...
– user3482749
Dec 4 '18 at 15:37










1 Answer
1






active

oldest

votes


















1














On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $varnothing$ to be open.



For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = varnothing$. This is, of course, open.






share|cite|improve this answer





















  • but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
    – maruto
    Dec 4 '18 at 16:18










  • No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
    – Randall
    Dec 4 '18 at 16:20








  • 1




    thank you Randall!
    – maruto
    Dec 4 '18 at 16:24






  • 1




    @ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
    – Randall
    Dec 4 '18 at 16:59








  • 1




    @Randall Thanks for the clarification.
    – Thomas Shelby
    Dec 4 '18 at 17:09











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $varnothing$ to be open.



For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = varnothing$. This is, of course, open.






share|cite|improve this answer





















  • but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
    – maruto
    Dec 4 '18 at 16:18










  • No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
    – Randall
    Dec 4 '18 at 16:20








  • 1




    thank you Randall!
    – maruto
    Dec 4 '18 at 16:24






  • 1




    @ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
    – Randall
    Dec 4 '18 at 16:59








  • 1




    @Randall Thanks for the clarification.
    – Thomas Shelby
    Dec 4 '18 at 17:09
















1














On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $varnothing$ to be open.



For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = varnothing$. This is, of course, open.






share|cite|improve this answer





















  • but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
    – maruto
    Dec 4 '18 at 16:18










  • No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
    – Randall
    Dec 4 '18 at 16:20








  • 1




    thank you Randall!
    – maruto
    Dec 4 '18 at 16:24






  • 1




    @ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
    – Randall
    Dec 4 '18 at 16:59








  • 1




    @Randall Thanks for the clarification.
    – Thomas Shelby
    Dec 4 '18 at 17:09














1












1








1






On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $varnothing$ to be open.



For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = varnothing$. This is, of course, open.






share|cite|improve this answer












On the plus side, you are thinking correctly about the preimage. This is, roughly, why the definition of a topology requires $varnothing$ to be open.



For your case (2), you have that $(a,b)$ consists of purely negative numbers. Hence, as you suspected, $f^{-1}(a,b) = varnothing$. This is, of course, open.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 16:05









Randall

9,14611129




9,14611129












  • but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
    – maruto
    Dec 4 '18 at 16:18










  • No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
    – Randall
    Dec 4 '18 at 16:20








  • 1




    thank you Randall!
    – maruto
    Dec 4 '18 at 16:24






  • 1




    @ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
    – Randall
    Dec 4 '18 at 16:59








  • 1




    @Randall Thanks for the clarification.
    – Thomas Shelby
    Dec 4 '18 at 17:09


















  • but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
    – maruto
    Dec 4 '18 at 16:18










  • No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
    – Randall
    Dec 4 '18 at 16:20








  • 1




    thank you Randall!
    – maruto
    Dec 4 '18 at 16:24






  • 1




    @ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
    – Randall
    Dec 4 '18 at 16:59








  • 1




    @Randall Thanks for the clarification.
    – Thomas Shelby
    Dec 4 '18 at 17:09
















but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
– maruto
Dec 4 '18 at 16:18




but does equation 2 of only having the empty set as preimage not contradict that $x^2$ is continuous? this is my fundamental issue and point of confusion
– maruto
Dec 4 '18 at 16:18












No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
– Randall
Dec 4 '18 at 16:20






No. The definition of $f: X to Y$ being continuous is that the inverse image of every open set should be open. In the event that your open set is like one in case (2), its preimage is STILL open since it is empty.
– Randall
Dec 4 '18 at 16:20






1




1




thank you Randall!
– maruto
Dec 4 '18 at 16:24




thank you Randall!
– maruto
Dec 4 '18 at 16:24




1




1




@ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
– Randall
Dec 4 '18 at 16:59






@ThomasShelby No. For example, the preimage of $(-1,2)$ is $(-sqrt{2}, sqrt{2})$, which is open.
– Randall
Dec 4 '18 at 16:59






1




1




@Randall Thanks for the clarification.
– Thomas Shelby
Dec 4 '18 at 17:09




@Randall Thanks for the clarification.
– Thomas Shelby
Dec 4 '18 at 17:09


















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