Classify the sets below as
Note: for the purpose of classification, we’ll consider \mathbb{C} to be the “ambient space” of all these sets, meaning that when we consider each set, we’ll think of it being “surrounded” by \mathbb{C}, rather than being isolated.
\mathbb{C}
For sure \mathbb{C} is open since for any point we can find an open disc centered on the point that is a subset of \mathbb{C}. Also, \mathbb{C} doesn’t have any boundary points, so the (empty) set of boundary points is contained within \mathbb{C}, which implies \mathbb{C} is closed.
We surely can’t contain all of \mathbb{C} is an open disc, so \mathbb{C} is unbounded.
Intuitively, \mathbb{C} is connected since we can’t find any two non-empty disjoint sets whose union is \mathbb{C}
\mathbb{R}
If \mathbb{C} is the ambient space of \mathbb{R}, then \mathbb{R} is closed since all of the boundary points of \mathbb{R} are elements of \mathbb{R}.
We can’t contain \mathbb{R} in an open disc, so it is unbounded.
\mathbb{R} is connected since we can’t find any two non-empty disjoint sets whose union is \mathbb{R}.
\mathbb{N}
If \mathbb{C} is the ambient space of \mathbb{N}, then \mathbb{N} is closed since all of the boundary points of \mathbb{N} are elements of \mathbb{N} (or better yet, because \mathbb{N}^\complement is open).
\mathbb{N} is unbounded since we can’t contain it in an open disc.
\mathbb{N} is disconnected since the union of the even naturals and the odd naturals forms the naturals.
\{z\in\mathbb{C}: \text{Re}(z)>0\}
This guy is open, unbounded, and connected.
\{z\in\mathbb{C}: \text{Re}(z)>0 \text{ and Im}(z) \geq 0 \}
This set is not open since those elements of the set with imaginary components equal to 0 are not interior points. This set is not closed since those complex numbers with real components equal to 0 are boundary points of the set but not members of the set.
This set is unbounded.
This set is connected.
\{z\in\mathbb{C}: z\bar{z}\leq 4\}=\{z\in\mathbb{C}: |z|^2\leq 4\}
This set is closed since its boundary points, \partial(\{z\in\mathbb{C}: z\bar{z}\leq 4\})
= \{z\in \mathbb{C}: |z| = 2\}, are all members of the set.
This set is bounded since the disc of radius 3 centered at 0 contains the set.
Lastly, this set is connected.
I agree. Can you identify the set precisely?
Is this the set?
(based on Filliam_wisher and Dr. McClure’s work)
where,
refers to the order of z and its accumulation point at
where,