Let's analyze each of the options and determine if they form open sets within the real line $\mathbb{R}$.

The set $A = (-\infty, 0]$ is a subset of the real line, which includes all real numbers less than or equal to 0. This set is not open because it contains its boundary point, 0.

Option A: $A \cap [0, 1]$

This is the intersection of two sets: $A = (-\infty, 0]$ and the closed interval $[0, 1]$. The intersection of these two sets is simply $[0]$, since $0$ is the only point that belongs to both $A$ and $[0, 1]$.

• $[0]$ is a single point, and single points are not open in $\mathbb{R}$.
• Therefore, A is not an open set in this case.

Option B: $A \cap (-\infty, -1)$

This is the intersection of the set $A = (-\infty, 0]$ with the open interval $(- \infty, -1)$. The intersection of these two sets is the interval $(- \infty, -1)$, since the set $(- \infty, -1)$ consists of all real numbers strictly less than $-1$, and these numbers are also in $A$.

• $(- \infty, -1)$ is an open set in $\mathbb{R}$ because it does not include any boundary points (it has no points of closure in its interval).
• Therefore, B is an open set.

Option C: $A \cup \{1/2\}$

This is the union of the set $A = (-\infty, 0]$ with the single point $\{1/2\}$. The set $A$ is already non-open as it contains the boundary point $0$. Adding a single point $\{1/2\}$ does not make the set open, since the union of a non-open set with a point does not create an open set.

• $\{1/2\}$ is a single point, and adding a single point to a non-open set does not result in an open set.
• Therefore, C is not an open set.

Option D: $A \cup (-1, 1)$

This is the union of the set $A = (-\infty, 0]$ with the open interval $(-1, 1)$. The union of these two sets will result in a set that includes all numbers less than or equal to 0 from $A$ and all the numbers strictly between $-1$ and $1$ from $(-1, 1)$. The resulting union is the set $(- \infty, 1)$, which is the open interval from $- \infty$ to $1$.

• $(- \infty, 1)$ is an open set because it does not include its boundary points, including 1.
• Therefore, D is an open set.

Option E: $A \cup (0.1, 1)$

This is the union of the set $A = (-\infty, 0]$ with the open interval $(0.1, 1)$. The set $A$ already includes all real numbers less than or equal to 0, and the interval $(0.1, 1)$ is a set of numbers strictly between 0.1 and 1. Thus, the union of these two sets is the set $(-\infty, 0] \cup (0.1, 1)$, which contains two disjoint parts: the set $(-\infty, 0]$ and the set $(0.1, 1)$.

• The set $(- \infty, 0]$ is not open, and adding the interval $(0.1, 1)$ does not make the union open, as the set $(- \infty, 0]$ includes the boundary point 0, which prevents the whole set from being open.
• Therefore, E is not an open set.

Final Answers:

• B: $A \cap (-\infty, -1)$ is an open set.
• D: $A \cup (-1, 1)$ is an open set.

Additional Questions:

1. What does it mean for a set to be open in $\mathbb{R}$?
2. Why are single points in $\mathbb{R}$ not open sets?
3. How do boundary points affect whether a set is open or not?
4. What is the difference between open and closed intervals?
5. Can the union of two non-open sets ever be open? Why or why not?

Tip:

Remember that a set is open if it does not include its boundary points and for every point in the set, there exists a small neighborhood around it that is entirely contained within the set.