1
$\begingroup$

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?

  • 0
    The usage of $][$ to denote the open side of an interval is not uncommon. It is used in the US as well. Marsden's "Elementary classical Analysis" is an example.2012-12-16

1 Answers 1

2

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.

  • 1
    $A$ only contains real numbers; their imaginary part is zero, which is not the case for $3+i\varepsilon/2$.2012-12-16