f is a continuous function from [0,1] to [0,1]

Why is $\displaystyle A=\{x\in[0,1]:f(x)\geq x\}$ closed?

Printable View

- Dec 7th 2010, 03:06 AMbram kierkelsClosed subset
f is a continuous function from [0,1] to [0,1]

Why is $\displaystyle A=\{x\in[0,1]:f(x)\geq x\}$ closed? - Dec 7th 2010, 03:49 AMDrSteve
Let the sequence $\displaystyle x_n$ converge to $\displaystyle x$ with $\displaystyle x_n\in A$ for all $\displaystyle n$. Then $\displaystyle f(x) = f(\lim x_n) = \lim f(x_n)\geq \lim x_n = x$. So $\displaystyle x\in A$ and $\displaystyle A$ is closed.

- Dec 7th 2010, 05:39 AMDrexel28
Or, note that $\displaystyle A=\left\{x\in[,1]:f(x)-x\geqslant 0\right\}$ and thus if $\displaystyle g(x)=f(x)-x$ then $\displaystyle g$ is (trivially) continuous and $\displaystyle A=g^{-1}\left([0,\infty)\right)$. In fact, although harder to prove it's true that this is the case in any order topology.