Recent content by iamjign

sorry i mixed up the symbol it should be the reverse. and h^{-1}{\{0\}} \subset R .

because {0} is closed in R and f is a function bijected R back to R. therefore, h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}\supseteq {\{0\}} \supset R?

Homework Statement assume h: R->R is continuous on R and let K={x: h(x)=0}. Show that K is a closed set. Homework Equations The Attempt at a Solution since we know h is continuous and h(x)=0. therefore, we know there is a epsilon neighborhood such that x belongs to preimage...