# If $V$ is a neighbourhood of $a$ having $E= D \cap V$ , show that $a \in E'$

I'm having a little problem in understanding this question...

Let $f \colon D \rightarrow \mathbb{R}$ with $D \subset \mathbb{R}$ and $a \in D'$. If $V$ is a neighbourhood of $a$ and $E= D \cap V$, show that $a \in E'$.

The exercise also asks to show that if $f|_E$ has a limit at the point $a$ then $f$ also has a limit at the point $a$ .

More specifically, I don't understand how to analyze the restricted function...

Thanks the attention!

(This is an exercise of my real analysis list of exercises. I spent a good time trying to do it, but I'm stuck... If anyone could give me a light... I'm studying for my final exam...)

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What is it you mean by $D'$? Is it the closure of $D$, the boundary of $D$, or the derived set of $D$ (set of all limit points)? – Arturo Magidin Jun 27 '12 at 18:29
@ArturoMagidin It is the derived set of D,the set of all limit points. – HipsterMathematician Jun 27 '12 at 18:30
The OP says "limit in the point" so I was assuming derived set. It would be better to make this clear before launching into the problem, though. I also don't see what $f$ has to do with anything. This seems to just be a question about $D$ and $V$. – Dylan Moreland Jun 27 '12 at 18:31
In the first question ,the question itself,no it doesn´t have to do with anything.But ,the exercise wants us to prove that if f|E has a limit in the point a then f also has a limit in the point a. – HipsterMathematician Jun 27 '12 at 18:35
@MeAndMath It might be better to write "at the point $a$". – Dylan Moreland Jun 27 '12 at 18:39

Because $a\in D'$, that means that every neighborhood of $a$ must contain points of $D$ other than $a$. You want to prove that every neighborhood $U$ of $a$ must contain points of $E\cap V$ other than $a$ itself; let $U$ be a neighborhood of $a$. Then $U\cap V$ is also a neighborhood of $a$, so we know that $(U\cap V)\cap D-\{a\}$ is not empty. But $(U\cap V)\cap D = U\cap (V\cap D) = U\cap E$. Hence...
For the second part, let us suppose that $f|_E$ has a limit at $a$, say $r$. This means that for every open set $W$ that contains $r$, there exists an open set $U$ that contains $a$ and such that $f((U\cap E)-\{a\})\subseteq W$. To show that $f$ has a limit at $a$, we need to show that there exists an open set $\mathscr{O}$ that contains $a$ and such that $f((\mathscr{O}\cap D)-\{a\})\subseteq W$.
Fix $W$, an open set containing $r$, and let $U$ be an open set that contains $a$ and such that $f((U\cap E)-\{a\})\subseteq W$. Now let $\mathscr{O} = U\cap \mathrm{int}(V)$, the interior of $V$. This is an open set that contains $a$ (since $U$ is open and contains $a$, and $V$ is a neighborhood of $a$ so its interior contains $a$). Will that work?
@MeAndMath: What you wrote does not even parse, so no. Writing a mathematical argument is not cooking pasta: you don't throw a lot of words at the wall and hope something sticks. The sentences have to make sense, let alone be logical. Also, "there are points different from $a$ in $E$" does not follow from "$a$ belongs to $E$". Why is $E$ not a singleton? And how are you concluding that $U\cap E$ contains points other than $a$? – Arturo Magidin Jun 27 '12 at 19:01