Euclidean topology: preimage of $(a, b)$ with $a < b leq 0$
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
asked Dec 4 '18 at 15:36
maruto
33
add a comment |
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
asked Dec 4 '18 at 15:36
maruto
33
1
It couldn't. Therefore...
– user3482749
Dec 4 '18 at 15:37
add a comment |
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
asked Dec 4 '18 at 15:36
maruto
33
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
asked Dec 4 '18 at 15:36
maruto
33
asked Dec 4 '18 at 15:36
maruto
33
edited Dec 4 '18 at 15:49
asked Dec 4 '18 at 15:36
maruto
33
asked Dec 4 '18 at 15:36
maruto
33
asked Dec 4 '18 at 15:36
maruto
33
33
1
It couldn't. Therefore...
– user3482749
Dec 4 '18 at 15:37
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 4 '18 at 16:05
Randall
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
|
show 4 more comments
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1 Answer
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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.
answered Dec 4 '18 at 16:05
Randall
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
|
show 4 more comments
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.
answered Dec 4 '18 at 16:05
Randall
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
|
show 4 more comments
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.
answered Dec 4 '18 at 16:05
Randall
9,14611129
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.
answered Dec 4 '18 at 16:05
Randall
9,14611129
answered Dec 4 '18 at 16:05
Randall
9,14611129
answered Dec 4 '18 at 16:05
Randall
9,14611129
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
|
show 4 more comments
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
|
show 4 more comments
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1
It couldn't. Therefore...
– user3482749
Dec 4 '18 at 15:37