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Let $$ F : \mathbb{R^2} \to \mathbb{R}$$

with $\mathbb{X} = \begin{pmatrix} x \\ y \end{pmatrix}$

$$ F(\mathbb{X)}=\left\{ \begin{array}{ll} \text{ 0 when x $\in \mathbb{Q}$ and y $\in \mathbb{Q}$}\\ \text{1 else }\\ \ \end{array} \right. $$

My question is that equivalent to

$ F(\mathbb{X)}=\left\{ \begin{array}{ll} \text{ 0 when x $\in \mathbb{Q}$ and y $\in \mathbb{Q}$}\\ \text{1 when x $\in \mathbb{R\backslash Q}$ and y $\in \mathbb{Q}$ }\\ \text{1 when x $\in \mathbb{Q}$ and y $\in \mathbb{R\backslash Q}$ }\\ \text{1 when x $\in \mathbb{R\backslash Q}$ and y $\in \mathbb{R\backslash Q}$ }\\ \end{array} \right. $

if so then how to proof that F is not continuous in $\mathbb{R^2}$

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    1) Your formulation is equivalent. 2) In order to show "not continuous", you just need one "counterexample". Since this is about numbers being rational/irrational, maybe come up with rationals numbers converging to an irrational, or vice versa. 3) If you have a problem with the fact this is about a 2-dimensional domain, try to formulate and solve the problem for a 1-dimensional domain ($\mathbb{R} \to \mathbb{R}$) first – 2017-01-31
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    Plz tell me if thqt correct i will handel the case when f(x)= 1 all those 3 cases can be resovled in the same way let $X_{0}$= (a,b) with a $\in \mathbb{R\backslashQ}$ and b$ \in \mathbb{Q}$ – 2017-01-31
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    so lets say that f is continuious in $X_{0}$ then exists a $ \epsilon$ and $\delta$ so that : let $\epsilon = 0,5$ $\abs{f(x) -f($X_{0}$)}< \epsilon and \norm{\abs{x}-\abs{$X_{0}$}}< \delta $ $\abs{x1 -a} < \delta$ $a-\delta – 2017-01-31
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    It seems you have the right idea, but it is hard to tell with all the LaTeX things going wrong. – 2017-02-01
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    how can i write absolute value and norme vecteur value here ? – 2017-02-01
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    absolute value: $\left| x-y\right|$ is made by \left| x-y\right| norm: $\Vert x\Vert$ is made by \Vert x\Vert – 2017-02-01
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    Plz tell me if that correct, i will handel the case when f(x)= 1 all those 3 cases can be resovled in the same way let $X_{0}$= (a,b) with$ a \in \mathbb{R\textbackslash{Q}}$ and b$\in \mathbb{Q }$ – 2017-02-01
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    so let s say that F is cotinuious in $X_{0}$ then exists an $\epsilon$ and a $\delta$ so that : let $\epsilon = \frac{1}{2}$ and $\left| f(X)-f(X_{0})\right| < \epsilon$ and$ \vert \left|X\right| - \left|X_{0}\right| \vert < \epsilon$ – 2017-02-01
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    $\left|x1 -a\right| < \delta$ $\Rightarrow (a- \delta) $ (a- \delta) and (a+\delta) $are $\in \mathbb{R}$ then can be x1 $\in \mathbb{Q}$ , the same to x2 then will X =(x1,x2)$ \in \mathbb{Q^2}$ $\Rightarrow \left|f(x) -f(X_{0})\right| =\left| 0 -1\right| <0.5 \rightarrow $ that is not possible$ \rightarrow$ f is not in $X_{0}$ Continuious for all $X_{0} \in \mathbb{(R\Q*Q)}$ is that correct ? – 2017-02-01

1 Answers 1

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(If you are familiar with Connectedness)

Suppose $F$ was continuous, then image of connected set would be connected.

Consider the connected subset $[0,1] \times [0,1]$ of $\Bbb R^2$. Then $F([0,1] \times [0,1])=\{0,1\}$ and $\{0,1\}$ is not a connected subset of $\Bbb R$.

EDIT: In fact $\Bbb R^2$ is connected. But the image $F(\Bbb R^2)=\{0,1\}$ is not.