Is the Subset S Open, Closed, or Neither in Various Complex Spaces?

[Oxymoron]

Question 1

Let $\mathcal{H} = \mathbb{C}^k$, where $\mathcal{H}$ is a Hilbert space. Then let

$$S = \left\{x : \sum_{i=1}^{k} |x_i| \leq 1 \right\}$$

be a subset of $\mathcal{H}$. Is the subset $S$ open, closed or neither?



Question 2

Let $\mathcal{H} = \mathbb{C}$. Then let

$$S = \left\{\frac{1}{n} : n\in \mathbb{N}\right\}$$

be a subset of $\mathcal{H}$. Is the subset $S$ open, closed or neither?



Question 3

Let $\mathcal{H} = \mathbb{C}^2$. Then let

$$S = \left\{(z,0) : z\in \mathbb{C}\right\}$$

be a subset of $\mathcal{H}$. Is the subset $S$ open, closed or neither?



Question 4

Let $\mathcal{H} = l^2$. Then let

$$S = \left\{x : \sum_{i=1}^{\infty} |x_i|^2 < 1\right\}$$

be a subset of $\mathcal{H}$. Is the subset $S$ open, closed or neither?



Question 5

Let $\mathcal{H} = L^2([0,1])$. Then let

$$S = \left\{f : f(t) \neq 0 \, \forall \, t \in [0,1]\right\}$$

be a subset of $\mathcal{H}$. Is the subset $S$ open, closed or neither?

 

[Oxymoron]

Solution 1

Intuitively, this set is closed because it contains its own boundary.

 

[Oxymoron]

Solution 2

We can write out the subset as

$$S=\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}$$

In set notation...

$$S = (0,1]$$

That is, S is neither open nor closed.

 

[Oxymoron]

Solution 3

Since $(z,0) = 0 \, \forall \, z \in \mathbb{C}$, then S is closed since the complement is open. ie since

$$\mathbb{C}^2 \backslash \left(S = \{0\}\right)$$ is open.

 

[HallsofIvy]

Solution 2

We can write out the subset as

$$S=\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}$$

In set notation...

$$S = (0,1]$$

That is, S is neither open nor closed.

? I would interpret (0, 1] as the set of all real numbers between 0 and 1 (including 1 but not 0) not S.

Of course, it is true that the set is neither open nor closed. It is not open because a neighborhood of 1/n, a disk in the complex plane centered on 1/n will contain numbers not in the set. It is not closed because the sequence has limit point 0 which is not in the set. (If 0 were included, the set would be closed.)

 

[HallsofIvy]

Solution 3

Since $(z,0) = 0 \, \forall \, z \in \mathbb{C}$, then S is closed since the complement is open. ie since

$$\mathbb{C}^2 \backslash \left(S = \{0\}\right)$$ is open.

I may be misunderstanding your notation. It is true that S is closed because the complement of S is open.
But I don't understand your saying (z, 0)= 0 . (z, 0) (for all complex z) is topologically equivalent to the complex plane in the same way that the x-axis, y= 0 (all points (x, 0)), is topologically equivalent to the real line.

And I really don't understand what you mean by $$\mathbb{C}^2 \backslash \left(S = \{0\}\right)$$. What could S= {0} mean?

 

[Oxymoron]

? I would interpret (0, 1] as the set of all real numbers between 0 and 1 (including 1 but not 0) not S.

Ha. I don't know what I was talking about ? You are right of course, and what you wrote is exactly what I was thinking...

I may be misunderstanding your notation. It is true that S is closed because the complement of S is open. But I don't understand your saying (z, 0)= 0 . (z, 0) (for all complex z) is topologically equivalent to the complex plane in the same way that the x-axis, y= 0 (all points (x, 0)), is topologically equivalent to the real line.

I see. I wasnt sure what was meant by (z,0) - but from what you wrote I would guess that what the question means. I suppose I had a lucky guess then!?