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The question asks,

Find a function $g \colon A \to \mathbb{R}$ and a set $B \subset A$ such that $g|B$ is continuous but $g$ is continuous at no point of $A$.

My idea is this:

Let $ A = \{\frac{1}{n}\, | \, n= 1, 2, \ldots \} $ and $ g\colon A \to \mathbb{R} $ be defined by $ g = \begin{cases} 0 : x=0\\ x : x \ne 0 \end{cases} $. We see that $ g $ is continuous on $ B = \{0\} $, but not for any point in $ A $.

First, does it even make sense to define $B$ as a finite set with one element? How can values of $x$ "approach" $0$ if there are no other values in $B$? Do the values that approach $0$ come from $A$ (for which $0$ is an accumulation point)? More generally, when we restrict the domain of a function as we have here, do we still look at the ambient set for values when determining limits, or are we also restricting ourselves to the subset?

Second, related to my first question is the second part of the problem. Can we talk about $\lim_{x \to \frac{1}{4}}g$ since $\frac{1}{4}$ is an "isolated" point? That is, we may not be able to find $x$-values near $\frac{1}{4}$.

Third, could this example fix the above issues?

Let $ A = (0,1) $, $ B = \mathbb{Q}\cap (0,1) $ and $ g\colon A \to \mathbb{R} $ be defined by $ g = \begin{cases} x : x \in \mathbb{Q}\\ 0 : x \notin \mathbb{Q} \end{cases} $.

Kevin Sheng
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    $B$ isn't a subset of $A$... – Shane Nov 05 '14 at 22:17
  • $B$ needs to ruin the continuity of all points of $A$, which makes me think $B$ should be dense in $A$. – Austin Mohr Nov 05 '14 at 22:18
  • $A$ is a discreet set. So any function $f:A\to \mathbb{R}$ is continuous. – mfl Nov 05 '14 at 22:18
  • @Shane, oops... you're certainly correct. Let's make the change that $A_1 = A \cup {0}$. – Kevin Sheng Nov 05 '14 at 22:21
  • @KevinSheng Then you have a problem that $g$ is continuos at some point of $A$. Literally, you couls pick just any nonwhere continous function $g$ and let $B$ an arbitrary finte subset of $A$. – Hagen von Eitzen Nov 05 '14 at 22:22
  • And @HagenvonEitzen's example includes the easy example where $B$ is a singleton set, or even the empty set. So the most difficult part is coming up with a nowhere continuous function. – Jeppe Stig Nielsen Nov 05 '14 at 22:31

2 Answers2

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Define $g: R \to R$ by $g(x) = 1$ if $x$ is irrational, zero otherwise.

Then $g|\mathbb Q$ is continuous (indeed, constant).

John Hughes
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Note that any function on a discrete set is continuous with respect to the induced topology on the set, so $g$ is in fact continuous on $A$. A canonical example to what you seek is indeed what you gave last.

Another example is $$f(x) =\cases{\min\{n : nx \in \mathbb Z\} & $x\in\mathbb Q$\\0 & $x\notin\mathbb Q$}$$ Here $f|_{\mathbb R\setminus\mathbb Q}$ is the continuous restriction. $f$ itself isn't even bounded.

AlexR
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