To determine which subsets are open, recall that a set is open if, for every point in the set, there exists an open interval around that point that is entirely contained within the set. Now let’s analyze each option based on the subset $A = (-\infty, 0]$.

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1. $A \cap [0, 1]$

The set $A = (-\infty, 0]$ intersects $[0, 1]$ at $\{0\}$. Since $\{0\}$ is a single point and no interval around $0$ lies entirely within $\{0\}$, this set is not open.

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2. $A \cap (-\infty, -1)$

The set $A \cap (-\infty, -1)$ is the intersection of $(-\infty, 0]$ and $(-\infty, -1)$, which gives $(-\infty, -1)$. This is an open interval, so this set is open.

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3. $A \cup \{1/2\}$

The union $A \cup \{1/2\}$ adds the point $1/2$ to $A$. Since a single point, like $\{1/2\}$, is not open, and adding it does not create an interval around $1/2$, this set is not open.

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4. $A \cup (-1, 1)$

The union $A \cup (-1, 1)$ creates a set that includes the interval $(-1, 1)$ and all of $A = (-\infty, 0]$. While $(-1, 1)$ is open, the set as a whole includes the boundary point $0$ from $A$, which is not an interior point (no interval around $0$ lies entirely in the set). Therefore, this set is not open.

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5. $A \cup (0.1, 1)$

The union $A \cup (0.1, 1)$ creates a set that includes the interval $(0.1, 1)$ and $A = (-\infty, 0]$. The presence of $0.1$, $1$, and points in $(0, 0.1]$ means $A \cup (0.1, 1)$ is not open, as the boundary points of $A$ and $(0.1, 1)$ break openness.

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Conclusion

The open sets are:

• $A \cap (-\infty, -1)$.

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Would you like further clarifications or details about open and closed sets? Here are 5 related questions you may find useful:

1. What does it mean for a set to be closed instead of open?
2. Can a set be both open and closed (a "clopen" set)?
3. How does the concept of openness change in higher dimensions?
4. Why is $[0, 1]$ not an open set in the real line?
5. How do unions and intersections of open sets behave?

Tip: To check if a set is open, always test whether every point in the set has a neighborhood entirely contained within the set.