I'm not sure where to start or how to even attempt this problem. I need to find a function for $A= \{x \in \mathbb R |0\leq x \leq 1 \}\cup {2}$.
Help please!
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Uh-A={2} and A= [0,1] ?Does that make sense? You must have made a mistake. – Mathemagician1234 Mar 29 '15 at 22:00
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I edited your question-is this what you meant? – Mathemagician1234 Mar 29 '15 at 22:30
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Consider the following mapping where A =[0,1}$\cup${2} and f:R$\rightarrow$R defined by:If $0\leq x \leq 1$ or x=2, then f(x) =1 iff x is rational and 0 if x is not rational. If $x\notin [0,1]$, then f(x) = x.It's easy to check f is discontinuous at every x in A= [0,1] and continuous everywhere else on the real line. Notice this function covers the case where x = 2 since 2 is rational i.e. 2 = $\frac{2}{1}$.
I'm not sure if this is the domain you wanted, your question isn't that clear. But this should give you the kind of function you're seeking and it's easily modified.
Mathemagician1234
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According to the title of your question..Although the function is NOT defined on your set $A$.
Consider Thomae's function...which is discontinuous at all rational points , i.e. in the set $\mathbb Q$..but is continuous in $\mathbb R \setminus \mathbb Q$.
Consider , $f:\Bbb R\to \Bbb R$ by $$f(x)=\begin{cases}1&\text{ if } x=0\\0 & \text{ for } x\text{ is irrational. }\\\frac{1}{q} & \text{ if } x=\frac{p}{q}\text{ where }p,q\text{ are relatively prime with }q\not =0\end{cases}$$
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if a seq of rational converges to zero its image seq also converges to zero.. – neelkanth Dec 16 '15 at 17:37
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1If $A\subset R$ then there exists $f:R\to R$ which is continuous at each $x\in R\backslash A$ and discontinuous at each $x\in A$ if and only if $A$ is an $F_{\sigma}$ set in $R$. – DanielWainfleet Dec 17 '15 at 03:53
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