Subset of $\mathbb{C}$ 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. 
I know that the set is closed as the complement is open. But I don't really know how to show this. And I also know that the set is connected. But I do not know how to show this.
Any help will be appreciated.
 A: can you describe these points and make a picture of your set? 
it helps a lot.
it is an "infinite triangle" - in $\mathbb{R}^2$ it is a positive (meaning that $x\ge 0$ and $y\ge 0$) part of plane between lines $y=0$ and $y=x$ (with both of them)
A: Let $A=\left\{z \in \mathbb{C} \setminus \{0\} | 0 \leq \arg(z) \leq\frac\pi4\right\}\cup\{0\}$ and the function $w:A\to[0,1]$ which $z=re^{it}\to\tan t$ is continuous.
A: It is easy to show that your set can also be described by
\begin{equation*}
A = \left \{ x + iy \mid x \ge 0 \text{ and } 0 \le y \le x \right \}
\end{equation*}
Let $ (z_n)_{n\in \mathbb{N}} $ be a converging sequence of complex numbers such that $z_n \in A$ for all $n\in \mathbb{N}$ and $z$ its limit. To show that $A$ is closed it suffices to show that $z\in A$. 
If we write $z_n = x_n +iy_n$ we have 
\begin{cases}
0 \le x_n \\
0 \le y_n \le x_n
\end{cases}
for all $n\in \mathbb{N}$. Because limits conserve inequalities (meaning that if $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ are two converging sequences of real numbers with $a_n \rightarrow a$ and $b_n \rightarrow b$ then $a_n \le b_n$ or $a_n < b_n$ for all $n\in \mathbb{N}$ implies $a\le b$) we have, if we write $z = x+iy$, 
\begin{cases}
0 \le x \\
0 \le y \le x
\end{cases}
which is equivalent to saying $z\in A$. 
To show that $A$ is connected it is easier to show the stronger statement $A$ is convex (if a set is convex then it is connected). Let $z_1 = a+ib$ and $z_2 = x+iy$ be two points of $A$. We need to show that if $t\in [0,1]$ then $z_t = tz_1+(1-t)z_2$ is also in $A$. We have 
$$
z_t = ta+(1-t)x + i(tb+(1-t)y)
$$
so we need
\begin{cases}
0 \le ta+(1-t)x \\
0 \le tb+(1-t)y \le ta+(1-t)x.
\end{cases}
The first two inequalities are trivially respected because the sum of two non-negative reals is also non-negative. We know that $b\le a$ and $y\le x$ therefore $tb+(1-t)y \le ta+(1-t)x$ which suffices to conclude that $A$ is convex.
A: Closed
Note that $A=\arg^{-1}([0,\frac{\pi}{4}])\cup\{0\}$ which is closed as a union of two closed sets, since $[0,\frac{\pi}{4}]$ is closed and $\arg$ is continuous.
This post shows some proof of arg(z) is continuous.
Note : these proofs show continuity for a cut on the ray $\{x\ge0,y=0\}$ and $\arg(z)\in[0,2\pi[$, but you can place the cut elsewhere for instance on the ray $\{x\le 0,y=0\}$ and have $\arg(z)\in[-\pi,\pi[$ so the cut is not bothering us when considering $[0,\frac{\pi}{4}]$.
Connected
Here the simplest method to show it is connected it to show that it is convex. For any two points in this infinite pie then the segment is also contained in the pie.
Let have $z_1=x_1+i.y_1$ and $z_2=x_2+i.y_2$ and $\lambda\in[0,1]$ 
Now what can we say about $z=x+i.y=\lambda.z_1+(1-\lambda).z_2$ ?
$\begin{cases} 
z_1\in A\Rightarrow \quad x_1\ge 0\qquad 0\le y_1\le x_1 \\
z_2\in A\Rightarrow \quad x_2\ge 0\qquad 0\le y_2\le x_2 
\end{cases}$
We conclude immediately 
$\begin{cases}
x=\lambda.x_1+(1-\lambda).x_2\ge 0 \\
y=\lambda.y_1+(1-\lambda).y_2\ge 0 \\
y=\lambda.y_1+(1-\lambda).y_2\le\lambda.x_1+(1-\lambda).x_2=x
\end{cases}$
So $z\in A$.
