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(1) Show, by example, that an infinite intersection of open sets in $\mathbb{C}$ need not be an open set in $\mathbb{C}$.

Consider $\bigcap_{i}^{\infty}A_i \subset \mathbb{C}$ for $A_i$ open. Each $A_i$ has the property that for each point $a$ in the set there is an $\epsilon>0$ such that $B_\epsilon(a)\subset A_i$...

(2) Show that every open set in $\mathbb{C}$ can be written as a countable union of open balls.

Let $A\subset\mathbb{C}$ be open. Define a rational open ball with radius $r\in\mathbb{Q}$ around a point $z_0\in\mathbb{C}$ in $\mathbb{C}$ as $B_r(z_0)$=$z\in =\mathbb{C}||z-z_0|$} < r. By definition of open sets $\mathbb{C}$, $\forall a\in A$, $\exists r>0$ such that $B_r(a)\subset A$. Thus given $B_r(a_1)$,$B_r(a_2)$,...,$B_r(a_n)$, these balls cover $A$ and since each contains a rational number they are countable.

(3) Show, by example, that there are open sets in $\mathbb{C}$ for which the open balls in (2) cannot be made pairwise disjoint.

Without knowing the exact proof for (2) I'm not sure about this one.

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For (1): have you tried? For (2): can you think of a replacement for $\mathbb Q$, that is, a dense countable subset of $\mathbb C$? –  Mariano Suárez-Alvarez Jan 24 '12 at 6:54
    
The proof I have in $\mathbb{R}$ relies on the $\epsilon-property$ which clearly doesn't hold in $\mathbb{C}$ which makes me doubtful. –  Emir Jan 24 '12 at 7:30
    
Since I don know what your proof for $\mathbb R$ looks like nor what the $\varepsilon$-property is, I can't help. But do not be doubtful: actually try to do it, and if it works, great. and if it doesn't, oh well: but «being doubtful» is not an good state to be in in these matters! Maybe if you expanded your question to show what you have tried we could help you with it. –  Mariano Suárez-Alvarez Jan 24 '12 at 7:44
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You should be advised that this website, while enriching and helpful, is not a proper substitute for a course or a book. –  Asaf Karagila Jan 24 '12 at 8:41
    
For (3) you can probably get away with the union of two overlapping open balls. (In the case of $\mathbb{R}$, the union of overlapping intervals just yields another interval, but in $\mathbb{C}$, the union of overlapping balls may not yield a ball.) –  Arthur Fischer Jan 24 '12 at 15:15

3 Answers 3

up vote 5 down vote accepted

A hint for (2): Call an open ball rational if the coordinates of its center and the radius are rational. Then from the definition of "open" (draw a figure!) conclude that any open set $\Omega\subset {\mathbb C}$ can be written as union of rational balls.

A sketch of proof for (3): Assume there is a connected component $\Omega'$ of the given open set $\Omega$ which is not an open ball. Then no covering $\cup_\iota B_\iota$ of $\Omega'$ with open balls $B_\iota\subset\Omega'$ contains a $B_\iota$ making up all of $\Omega'$. Consider one of these $B_\iota$'s. There is a point $z_0\in \Omega'\setminus B_\iota$, and as $\Omega'$ is arcwise connected there is an arc contained in $\Omega'$ connecting $z_0$ to the center $z_\iota$ of $B_\iota$. It follows that there is a point $z_1\in\partial B_\iota\cap\Omega'$. This point $z_1$ cannot be covered by some other open ball $B_{\iota'}$ which is disjoint from $B_\iota$.

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Thanks for the hint -- I've edited my answer with a new attempt. –  Emir Jan 24 '12 at 15:05
    
Dear Christian: Very nice answer! +1! –  Pierre-Yves Gaillard Jan 24 '12 at 18:10

I'll try and shed some light as to why we can cover every open set in $\mathbb{R}$ as a countable number of disjoint open intervals (Take uncountably many disjoint points as an example for why it can't be every set), and why this wouldn't extend to $\mathbb{C}$. What allows us to achieve such a covering is the fact that $\mathbb{R}$ is a complete ordered field, which leads us to the fact that all connected open sets in $\mathbb{R}$ are some form of intervals. We simply cover each connected component of our open set with the appropriate maximal interval, check that they are disjoint, and point at the rationals to establish countability.

However, we don't have a coherent order on $\mathbb{C}$ that agrees with the ordering on $\mathbb{R}$ and the standard metric on $\mathbb{C}$, meaning we can't characterize the connected sets on $\mathbb{C}$ nearly as easily. The devil is in the pairwise disjointness of our prescribed covering - try and think of a set that you think won't work geometrically, look at its connected components, and derive a contradiction. Remember, a countable union of open sets is open.

As for part 1, try and decode what you mean by the "$\epsilon$-property" and translate it to $\mathbb{C}$, or perhaps a more general Euclidean space at first. (A big hint would be to try and consider what the open sets of radius $\epsilon$ are in $\mathbb{C}$.) Try and look past your doubt! Explore as many ideas as you can, and always try and stretch the limits of techniques and concepts you understand to their maximum capacity, in order to adapt them to new situations or see where they break down. A full examination into why the natural ideas don't work can be more initially illuminating than trying to find a correct approach immediately.

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To generalize Christian's nice answer to $(3)$:

A connected space cannot be covered by a family of disjoint nonempty open subsets containing more than one member.

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