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I am to find the proper number from $x \in \{2,3,4\}$ for which this following set is a neighborhood in $\mathbb{R}$ or in $\mathbb{C}$, $$A:= \left] 1,4 \right[ \cap \left[ 2,5 \right]$$

Firstly, I don't really understand what a neighborhood is. But what I understand till now is this:
Some number $x$ lies in some set. and we choose some $\epsilon$ in this set and we say "the $\epsilon$-neighborhood of $x$ is the set which lies with $|\epsilon|$-radius around $x$". Am I okay what this?

So as for my problem: $A:= \{2,3,4\}$, right? Then how can I choose a number from $\mathbb{R}$ and from $\mathbb{C}$ which lies in my $x$-set so that $A$ will be a neighborhood?

What is difference between complex and reals in terms of neighborhood?

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Your neighborhood definition is close to being a typical definition. Change $\epsilon$ to a non-negative real number and drop the absolute value from "$\epsilon$-radius". Here's Rudin's definition from PMA: "A neighborhood of $p$ is a set $N_r(p)$ consisting of all point $q$ such that $d(p,q)<r$." Here $d(x,y)$ is the distance function in the metric space in question which maps the points $x$ and $y$ to a non-negative real number with certain restrictions. – Todd Wilcox Dec 16 '12 at 8:34
Oh okay. thanks – doniyor Dec 16 '12 at 8:36
On the real line, a neighborhood looks like an open segment. In the complex plane, a neighborhood looks like the interior of a circle (also open). I'm sorry to be totally ignorant, can you clarify what "]$1,4$[" means? – Todd Wilcox Dec 16 '12 at 8:37
@ToddWilcox ][ is an open interval, is that wrong latex ? – doniyor Dec 16 '12 at 8:39
I wouldn't say it's "wrong". In the USA I've only ever seen $(1,4)$ used to denote the open interval from $1$ to $4$. There's so much different notation in the world that I would never say any of it is wrong or right, just what people have seen before and what they haven't. So $A=[2,4)$? Why didn't the problem just say that? And if I'm reading that right, $[2,4)$ can never be a neighborhood using Rudin's definition because it is only half open. – Todd Wilcox Dec 16 '12 at 8:40
up vote 2 down vote accepted

A set $U$ is a neighborhood of a number $a$ in $\Bbb R$ or $\Bbb C$ if you can choose an $\varepsilon>0$ such that every number with distance $<\varepsilon$ to $a$ is contained in $U$.

The set $A$ should probably be $[2,4[$ since it should be a neighborhood in $\Bbb R$ or $\Bbb C$. Since it doesn't contain $4$, it can't be a neighborhood of that number. It also cannot be a neighborhood of $2$ since no matter how small $\varepsilon>0$ you choose, the number $2-\varepsilon/2$ is not an element of $A$. So the only possibility is that it's a neighborhood of $3$.

In $\Bbb R$ this is indeed true since every number with distance $<1$ from $3$ is contained in $A$. In $\Bbb C$ however, no matter how small $\varepsilon>0$ you pick, the number $3+i\varepsilon/2$ is not in $A$ and therefore $A$ is not a neighborhood of $3$ in $\Bbb C$. In fact, $A$ is not a neighborhood of any number in $\Bbb C$ since it's too "thin", i.e. it contains no open disks.

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Wow, thanks man, great explanation. Thanks for your time!! :) – doniyor Dec 16 '12 at 9:49
Why are you saying +e/2 always, and not e alone. 2 - e also not in A,right? – doniyor Dec 16 '12 at 10:45
I chose $a-\varepsilon/2$ because the distance from $a$ to $a-\varepsilon$ is equal to $\varepsilon$, but I need a number whose distance is strictly less than $\varepsilon$. In other words, in order to show that $A$ is not a neighborhood of $2$ I must show that no matter which $\varepsilon>0$ I choose, there is a number which is not in $A$ but whose distance to $2$ is strictly less than $\varepsilon$. That is, I must show that no $\varepsilon$ satisfies the definition of neighborhood. – nonpop Dec 16 '12 at 11:02
Okay, cool, now i got it. Sorry for dumbness. Can u pls explain why 3+ie/2 not in A is? – doniyor Dec 16 '12 at 11:19
$A$ only contains real numbers; their imaginary part is zero, which is not the case for $3+i\varepsilon/2$. – nonpop Dec 16 '12 at 11:26

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