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?
Hi Aqua,
If $\displaystyle B$ is bounded we can choose a natural number $\displaystyle N$ large enough so that $\displaystyle B\subseteq [-N, N].$ Since $\displaystyle [-N, N]$ is compact, $\displaystyle f([-N, N])$ is compact (because $\displaystyle f$ is continuous). Hence, $\displaystyle f([-N, N])$ is bounded. Can you see why $\displaystyle f([-N, N])$ being bounded implies $\displaystyle f(B)$ 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?