Determine whether or not the given set is open, connected, and simply-connected. (Select all that apply.) (x, y) | 16 ≤ x2 + y2 ≤ 25, y ≥ 0 open connected simply-connected

A set is closed if contains all the point in its boundaries. A set is open if it doesn't contain any of the points in its boundaries. In this set, all the points of the boundaries are included because it is using the less than or equal to and greater than or equal to define the set.

The set is connected if you can find a path inside the set to connect any two points of the set. If you make the graph of the set you would see the set covers this condition because the set hasn't any division.

The set is simply connected if you can draw a closed curve inside the set and in the interior of the curve there are only points of the set. In other words, if the set has holes is not simply connected. This set doesn't have holes, it's simply connected.

MrRoyal MrRoyal · Answer 2 · 2021-10-24T15:09:50+02:00

Set is the collection or groups of related entities.

The true statements are: connected and simply-connected

The given parameters are:

[tex]\mathbf{16 \le x^2 + y^2 \le 25}[/tex]

[tex]\mathbf{y \ge 0}[/tex]

Notice that, the inequality is less than or equal to.

When the inequality in a set is less than or equal to or greater than or equal to, then the set is closed (i.e. not open)

From the graph of [tex]\mathbf{16 \le x^2 + y^2 \le 25}[/tex] (see attachment), we can see that:

A pair of the corner points (i.e. (0,5), (5,0), (0,-5) and (-5,0)) are directly connected and there is no division between the corner points; this means that, the set is connected.
All the four corner points of the set can be connected without holes; this means that, the set is simply connected.

Hence, the true statements are: connected and simply-connected