Question

Prove that for every sets X and Y in Rn, if X ⊆ Y then cl(X) ⊆ cl(Y ).

Answer 1

The proof involves using the definition of closure in topology, the properties of limit points, and the subset relationship between X and Y to show cl(X) is contained within cl(Y).

Step-by-step explanation:

To prove that for every sets X and Y in Rn, if X ⊆ Y then cl(X) ⊆ cl(Y), let us first define what the closure of a set is. The closure of a set S, denoted cl(S), in a topological space is the smallest closed set containing S. This includes all the limit points of S, and possibly points of S itself.

Now, proceeding with the proof:

1. Assume X ⊆ Y.
2. Since X ⊆ Y, every limit point of X is also a limit point of Y, because every open neighborhood of a limit point of X will also intersect Y at points different than the limit point.
3. Since cl(X) is the smallest closed set containing all points of X and its limit points, and these are also limit points or points of Y, cl(X) is contained within any closed set that contains Y.
4. Therefore, cl(Y), which is the closure of Y containing all points of Y and its limit points, must also contain all points of cl(X).
5. Hence, cl(X) ⊆ cl(Y).

Therefore, it is proven that if X ⊆ Y in Rn, then cl(X) ⊆ cl(Y).