Question

Let S be a infinite set and let T be a finite subset of S. Prove that S − T is infinite.

Answer 1

@UnkleRhaukus

Answer 2

i dont know

Answer 3

@mukushla

Answer 4

\[(S)=(x _{1},x _{2},x _{3},...,x _{n-1},x _{n},x _{n+1},...)\] \[(T _{n})=(x _{1},x _{2},x _{3},...,x _{n-1},x _{n}), T _{n}\subset S\]then\[(S)-(T _{n})=(x _{1},x _{2},x _{3},...,x _{n-1},x _{n},x _{n+1},...)-(x _{1},x _{2},x _{3},...,x _{n-1},x _{n})\] \[(S)-(T _{n})=(x _{n+1},x _{n+2},...)\]

Answer 5

I understand, thank you.

Answer 6

I think sometimes in higher level mathematics we tend to over think things way too much.

Answer 7

I was thinking in an alternative proof in the line tukajo was probably thinking at first.

We assume
\[T\cup (S-T)=S, \ S \cap T=\phi \]
Then we could think in the number of elements, or cardinal,
\[Card (T)+Card(S-T)=Card(S)\]
As T is finite, for some M, Card(T)<M. As S is infinite, for some N, Card(S)>N. So, suppose S-T is finite, then Card(S-T)<P, for some P. Then,
\[Card(T)+Card(S-T)<M+P\Rightarrow Card(S)<M+P\]
But this is in contradiction with our assumption, Card(S)>N, for some N.

Answer 8

Yes John, that is a bit more geared towards what I had thought. Thank you, do you have time in a moment to assist in developing my thoughts on one more problem?