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Proving that a complex set in open/closed/neither and bounded/not bounded

I think $\{z\in C:|z| = |\operatorname{re}(z)| +|\operatorname{im}(z)|\}$ is closed. But I have no idea how to show it since you have to take an element of the set (which lies on the axes) and take a neighbourhood around that (not all of the ball is within the set so it's not open).

But when you do the complement, not all of everything outside of the axes includes the ball (as some of the ball is on the axes). This would make it neither open nor closed but I'm sure it's closed! Can someone help me please?

It's definitely not bounded because no closed ball can cover all of the axes as they go on to infinity and beyond. Right?

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    What exactly do you mean? $|\text{re}(z)| + |\text{im}(z)|$ is an expression, not a set.2012-11-04
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    Sorry, the subset is called X6, I have edited to make it more clear, I hope...2012-11-04
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    Neither is 'neither open nor closed.'2012-11-04

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The set $E=\{z\in C:|z|=|Re(z)|+|Im(z)|\}$ is precisely the union of the horizontal and vertical axes on the complex plane. This is closed but not open.

To see that it is closed, show that the complement is open: for any $z$ in the complement of $E$, let $r=\min(|Re(z)|,|Im(z)|)$. Then the ball with centre $z$ and radius $r$ lies in the complement of $E$.

To see that it is not open, consider the balls centred at $z=0$ with arbitrary radius. Any such ball certainly contains a point not lying on either of the axes.

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    Yes, but how would I show it is closed?2012-11-04
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    Okay, so if I take a ball around the origin with any radius, it will show that not all of the ball is on the axis so it's not open. Now for the complement... If I show the same ball (but for the complement) it shows that the ball contains a point which wouldn't lie on either of the axes, making it open. Hence the actual set is closed. Am i right here?2012-11-04
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    Well what I did was: Draw the axes, draw a ball around any point on the axes... I said that not all of the ball is on the set, so it's not open. I wanted to do the same example for the complement, but what confuses me is that part of the ball is ON the axis, so not all of the complement is open?2012-11-04
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    Hmm, my lecturer definitely said you have to first show it's not open, then use the complement to decide whether it was closed or neither open nor closed. If you could just elaborate on how to show the complement is open from your post, I would be very grateful! :P2012-11-04
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    If you could just elaborate on how to show the complement is open from your post, I would be very grateful! :P2012-11-04
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    Unfortunately not :(. I thought the ball touched the axes, like it did when proving that it wasn't open. Are we using taking a different ball now or...?2012-11-04
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    So can I clarify... When showing it's not open... If I take a ball on the axes and show that a point within that ball is clearly not on the axes... It's not open. Now I just have to show it is closed.2012-11-04
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    Ah, so I've done right for that bit!2012-11-04
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    Thank you! I should be able to do it from the answer you posted originally. Things are a bit less blurry now haha.2012-11-04
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If you mean the set $\{z\in\mathbb C : |z|=|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\}$ you might want to consider the continuous function $g(z)=|z|-|\mathrm{Re}(z)|-|\mathrm{Im}(z)|$ and remember how continuous functions behave with closed sets.

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    Yes, I mean that, but unfortunately, I don't remember how they behave.2012-11-04
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    The preimage of a closed set is closed, and your set is $g^{-1}A$ for some closed $A$.2012-11-04
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EDIT

Looks like the question has now changed.


I assume you are interested in the following set $X$. $$X = \{\left \vert \text{Real}(z) \right \vert + \left \vert \text{Imag}(z) \right \vert : z \in \mathbb{C}\}$$

HINT

Note that $X$ contains only non-negative real numbers. It is easy to show that (Why?) $$X = \mathbb{R}^+_0 = \mathbb{R}^+ \cup \{0\}$$

Now look at the complement of $\mathbb{R}^+_0$ in $\mathbb{C}$ i.e. $X^c = \mathbb{C} \backslash \mathbb{R}^+_0$. Can you show that it is open?

HINT

Consider $z \in X^c$ and $z$ is not purely real. Consider an open ball centered at $z$ and radius $$r = \dfrac{\left \vert \text{Imag(z)} \right \vert}2$$

For $z \in X^c$ and is purely real, then $z$ is negative, For this $z$, consider an open ball centered at $z$ and radius $$r = \dfrac{\left \vert \text{z} \right \vert}2$$

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    That's what I'm trying to figure out, I thought you could show the complement was open, but now I'm thinking you can't.2012-11-04
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    Yes, X only contains non-negative real numbers, but it can take values from anywhere on any axis as when the modulus is taken, it will be in X.2012-11-04
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    If I read her question correctly, that’s the wrong set.2012-11-04
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The function $|z|-|\operatorname{im}(z)|-|\operatorname{re}(z)|$ is continuous. The set in question is the inverse-image of the closed set $\{0\}$ under this function. The inverse-image of a closed set under a continuous function is closed.

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    Now I see that someone else posted this same comment.2012-11-04