Let $GL(d,\mathbb{R} )$ set of invertiable matrices in $\mathbb{R^d}\times \mathbb{R^d} $.

Consider $f:GL(d,\mathbb{R})\rightarrow \mathbb{R} $be upper semi continues. Let X and Y are two subset of $GL(d,\mathbb{R})$.

If $\{f(x) ; x\in X\} $be closed interval and $d(X, Y) <\delta$, is $\{f(y) ; y\in Y\} $ closed interval?

I think we have to use definition of upper semi continuous since for every $\epsilon$ there is $\delta$ such that $d(X, Y) <\delta$ then $f(X) <f(Y) +\epsilon $


1 Answer 1


This is not true in general.

Taking $d = 1$, we have $GL(1, \mathbb R) \simeq \mathbb R \setminus\{0\}$. $f(x) = x$ is continuous, therefore upper semi-continuous. Now $f([1,2]) = [1,2]$ is a closed interval. However for all $1 >\delta >0$, $f((1- \delta, 2+\delta)) = (1- \delta, 2+\delta)$ is not closed.

  • $\begingroup$ Thank you very much for your answer, What about connected set?I mean instead of closed interval we consider connected set, is it true? Or not? $\endgroup$
    – Adam
    Oct 2, 2018 at 7:59

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