Closed subset

**bram kierkels** · December 7th 2010, 03:06 AM

f is a continuous function from [0,1] to [0,1]
Why is $A=\{x\in[0,1]:f(x)\geq x\}$ closed?

**DrSteve** · December 7th 2010, 03:49 AM

Let the sequence $x_n$ converge to $x$ with $x_n\in A$ for all $n$ . Then $f(x) = f(\lim x_n) = \lim f(x_n)\geq \lim x_n = x$ . So $x\in A$ and $A$ is closed.

**Drexel28** · December 7th 2010, 05:39 AM

Originally Posted by bram kierkels

f is a continuous function from [0,1] to [0,1]
Why is $A=\{x\in[0,1]:f(x)\geq x\}$ closed?

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

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