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Consider the following subset in $\mathbb{C}$

$$\left\{z \in \mathbb{C} \setminus \{0\} | 0 \leq \arg(z) \leq\frac\pi4\right\}\cup\{0\}$$

Determine if it is open,closed or neither and if it is connected.

Any help will be appreciated.

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    Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-02-22
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    You know what the argument represents, right? It represents the angle made by the complex vector with the positive real axis. Now, I suggest you actually attempt to sketch this region. There is a way of describing it in words, succinctly enough that everything about it is revealed, including the fact that it is connected, and is closed. – 2017-02-22
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    Can i have any sort of head start on proving this @астонвіллаолофмэллбэрг – 2017-02-22
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    One way of starting for you is the following: To show that this sector is closed, prove that it's complement is open. Take a point $p$ in the complement, this will have argument say $\operatorname{arg} p = \phi$, where $\phi > \frac \pi 4$ or $\phi < 0$. Break it up into cases, and find a neighbourhood of the point (hint: The sign is $<$, not $\leq$, so there will always be a positive distance to the given set from any point outside it) that is contained in the complement. This will prove closure. – 2017-02-22
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    To show connectedness, try to show that it is the preimage of a connected set under a continuous function. – 2017-02-22
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    Is $\arg$ continuous? What do you know about closed sets and continuous functions? – 2017-02-22
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    I have understood the brief reasoning behind why it is closed and connected, however im still finding it difficult to set out a proof to justify. Any help please @астонвіллаолофмэллбэрг – 2017-02-22
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    @user60589 can you explain your questions. – 2017-02-22
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    For closeness, the preimage of a closed set is closed, a closed intervall is closed and the function $\arg$ is continuous. – 2017-02-22
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    And for connectedness, just show path connectedness by taking any point and take the obvious path to $0$. – 2017-02-22
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    @астонвіллаолофмэллбэрг Could you elaborate how you want to show connectedness? Isn't any closed subset of $\Bbb R^2$ the preimage of a point? – 2017-02-22
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    @астонвіллаолофмэллбэрг: you probably meant the *image*, not the preimage. – 2017-02-22
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    @JackD'Aurizio You are right, I meant image, not preimage. – 2017-02-23

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You may easily figure out from this diagram that the complement is an open set.

enter image description here

Connectedness follows from path-connectedness: in order to join $A$ and $B$ in your region, you may just consider the segments $AO$ and $BO$, entirely contained in your region.

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    where are the A and B regions? can you label it – 2017-02-22
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    @user407151: $A$ and $B$ are two generic points inside your angle. Diagram updated. – 2017-02-22