If i have a continuous function "f" from the reals to the reals and a bounded subset of real numbers "B," is f(B) always bounded?
I'm trying to find some counter examples, but do counter examples even exist?
If is bounded we can choose a natural number large enough so that Since is compact, is compact (because is continuous). Hence, is bounded. Can you see why being bounded implies must be bounded?
Let me know if this gets things on the right track. Good luck!
I know by the preservation of compact sets if f:R->R is continuous on R and if B is a subset of R is compact then f(B) is compact.
If B were to be closed i know f(B) is not always closed (ex if g(x)=1/(1+x^2), B=[0, inf ) then g(B)=(0,1] which is not closed).
So your saying since B is inside the compact interval [-N,N], then f(B) must be bounded as well?