If $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$, then their union $Y_1\cup....\cup Y_n$ are also compact.
Proof:
Assume that $Y_1,...Y_n$ are a finite collection of compact subsets of $X$
WTS that $Y_1\cup....\cup Y_n$ are compact
In order for $Y_1 \cup...\cup Y_n$ to be compact, their union must be closed and bounded
Lets first prove that the union of $Y_1\cup...\cup Y_n$ is closed
I will do this using a previous proposition which I learned
Prop: If $Y_1,...Y_n$ are are finite collection of of closed sets in X, then $Y_1\cup...\cup Y_n$ is also closed
(I have the proof of this proposition)
By our assumption $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$. And since they are compact, then they must be closed
Now I have to prove that the union of $Y_1\cup...\cup Y_n$ is bounded
By our assumption $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$. And since they are compact, then they must be bounded
Since $Y_1,...,Y_n$ are bounded subsets of X then by definition $\exists B(x,r)$ in X which contains $Y_1,...,Y_n$
If $\exists B(x,r)$ in X which contains $Y_1,...,Y_n$ then this means that this ball $B(x,r)$ will also contain $Y_1\cup...\cup Y_n$ (Intuitively, this seems true to me, but I am not sure if it is and how to show this)
Can I please get some help in proving this?