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Define $g:[0,1]\rightarrow\mathbb R$ by $g(x)=\sqrt{x}$ if $x$ is rational and $g(x)=0$ if x is irrational.

Prove that $g$ is continuous at $x=0$, but is not continuous at any other value of $x$.

I am really at a loss of where to even start my proof. I know that in order for $g$ to be continuous as any point (let's say point $p$) then for every $\varepsilon>0$ where $E\subset X$ there exists a $\delta>0$ such that $d_Y(g(x),g(p))<\varepsilon$ for all points $x\in E$ for which $d_X(x,p)<\delta$

I'm having trouble applying this definition.

blubberbrot
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  • Hint: the contrapositive of $\forall (\epsilon > 0)\exists(\delta > 0) : (|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon)$ is $\exists (\epsilon > 0) \forall (\delta > 0), (|x-y| < \delta, |f(x) - f(y)| \geq \epsilon).$ Let $x \in \mathbb{R}$. If $x \neq 0$, choose $\epsilon = \sqrt{x}/2 > 0$. Now no matter how small a $\delta$ you choose, there is always some $y$ close to $x$ such that... – snar Apr 21 '14 at 03:47

2 Answers2

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a) $g(x)$ is continuous at $x=0$:

Proof:

Let $x_n \in [0,1]$ such that $x_n \to 0$, then $g(x_n) = 0$ if $x_n$ is irrational and $g(x_n) = \sqrt{x_n} \leq x_n$ for rational $x_n \in [0, 1]$. Hence, $\lim_{n \to \infty} g(x_n) = 0 = g(0)$ for any sequence $x_n \to 0$, so $g$ is continuous at $x = 0$.

b) $g(x)$ is not continuous at any other point in $[0, 1]$.

Let $x \in (0,1]$. Then if $x$ is rational, $g(x) = \sqrt{x} > 0$. Now just choose a sequence of irrational number $y_n \to x$, then $\lim g(y_n) = 0 < g(x)$. A similar argument works if $x$ is irrational.

Mustafa Said
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TO prove $g(x)$ is discontinuous at every point in $[0,1]$ except at $x=0$, consider the function

$$ f(x) = \frac{1}{\sqrt{x}} $$

Notice $f$ is continuous at every point except at $x = 0$. Suppose for contradiction that $g$ is continuous at every point $[0,1]$ except at $x=0$, then the function

$$ \frac{g}{f} $$

must be continuous also. But

$$ \frac{g}{f} = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right. $$

which we know is discontinuous everywhere. Contradiction!