If S is a compact subset of R and T is a closed subset of S, then T is compact. Prove this using the definition of compactness.

It has been proved that T is compact

Step-by-step explanation:

To prove this using the definition of compactness, let's assume that T is

not compact. Now, if that be the case, an open cover of T will exist. Let's call this open cover "A". Now, this open cover will have no finite subcover.

Now, from the question, since T is closed, it’s complement R\T will be open.

Therefore, if we add the set R\T to the collection of sets A, then we'll have an open cover of R and also of S.

Due to the fact that S is compact, this

cover will have a finite sub - cover which we will call B.

Finally, either B itself or B\{R\T} would be a finite sub - cover of A. This is a contradiction.

Thus, it proves that T has to be compact if S is to be a compact subset of R and T is to be a closed subset of S.

step-by-step explanation:

goog be able to you with that maybe

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm