- #1
varygoode
- 45
- 0
[SOLVED] Continuity on a piece-wise function
Problem:
Suppose:
[tex] f(x)=\left\{\begin{array}{cc}x^2, &
x\in\mathbb{Q} \\ -x^2, & x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right [/tex]
At what points is [tex] f [/tex] continuous?
Relevant Questions:
This is in a classical analysis course, not a real analysis course. So we use the metric defininition of continuity:
A function [tex] f:X \rightarrow Y [/tex], with [tex] X, Y [/tex] metric spaces and the distance between [tex] x, x_0 \in X, y, y_0 \in Y [/tex]
denoted as [tex] d_X(x, x_0), d_Y(y, y_0) [/tex], respectively, is called [tex] continuous \ at \ a \ point \ x_0 [/tex] if, [tex] \forall \ \varepsilon > 0 \ \exists \ \delta > 0[/tex] such that:
So, I'm just assuming that our distances are regular distance in [tex] \mathbb{R} [/tex], as this assumption is normal in my class when both sets are over subsets of [tex] \mathbb{R} [/tex].
Now, my question is as follows: should there be four cases here? One with both [tex] x, x_0 [/tex] (as in the definition) in [tex] \mathbb{Q} [/tex], one with both in [tex] \mathbb{R}\setminus\mathbb{Q} [/tex], and two with one in each? Is there more, less? Am I looking at this the wrong way?
Solution Attempt:
I started doing it with the four situation case, and I was getting that it is continuous only when [tex] x, x_0 \in \mathbb{R}\setminus\mathbb{Q} [/tex]. But I may be thinking about this all wrong.
Any ideas/suggestions/corrections/praise?
Problem:
Suppose:
[tex] f(x)=\left\{\begin{array}{cc}x^2, &
x\in\mathbb{Q} \\ -x^2, & x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right [/tex]
At what points is [tex] f [/tex] continuous?
Relevant Questions:
This is in a classical analysis course, not a real analysis course. So we use the metric defininition of continuity:
A function [tex] f:X \rightarrow Y [/tex], with [tex] X, Y [/tex] metric spaces and the distance between [tex] x, x_0 \in X, y, y_0 \in Y [/tex]
denoted as [tex] d_X(x, x_0), d_Y(y, y_0) [/tex], respectively, is called [tex] continuous \ at \ a \ point \ x_0 [/tex] if, [tex] \forall \ \varepsilon > 0 \ \exists \ \delta > 0[/tex] such that:
[tex] d_X(x, x_0) \leq \delta \Rightarrow d_Y(f(x), f(x_0)) \leq \varepsilon [/tex]
So, I'm just assuming that our distances are regular distance in [tex] \mathbb{R} [/tex], as this assumption is normal in my class when both sets are over subsets of [tex] \mathbb{R} [/tex].
Now, my question is as follows: should there be four cases here? One with both [tex] x, x_0 [/tex] (as in the definition) in [tex] \mathbb{Q} [/tex], one with both in [tex] \mathbb{R}\setminus\mathbb{Q} [/tex], and two with one in each? Is there more, less? Am I looking at this the wrong way?
Solution Attempt:
I started doing it with the four situation case, and I was getting that it is continuous only when [tex] x, x_0 \in \mathbb{R}\setminus\mathbb{Q} [/tex]. But I may be thinking about this all wrong.
Any ideas/suggestions/corrections/praise?