Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a locally-bounded set-valued mapping $f: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$ and the set-valued mapping $F: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$ defined as

$$ F(x) := \text{closure}(f(x)). $$

Question: is the mapping $F$ Outer SemiContinuous?

Note: definition of Outer SemiContinuity for a set-valued map.

A set-valued mapping $S: \mathbb{R}^n \rightrightarrows \mathbb{R}^m $ is outer semicontinuous at $\bar x$ if

$$ \limsup_{x \rightarrow \bar x} S(x) \subset S(\bar x) $$

or equivalently $\limsup_{x \rightarrow \bar x} S(x) = S(\bar x)$.

share|cite|improve this question
The definition of $F$ is not clear. What do you mean? – Jochen Sep 7 '12 at 12:17
Did you mean $f\colon \mathbb{R}^n \rightrightarrows \mathbb{R}^m$? Otherwise $\mathop{\mathrm{closure}} f(x)$ doesn't seem to make sense. – martini Sep 7 '12 at 12:51
Martini, you are right. I meant $f$ set-valued. Thanks. – Adam Sep 7 '12 at 13:58
up vote 1 down vote accepted

Counterexample in one dimension: $f(x)=0$ if $|x|\le1$ and $f(x)=1$ otherwise. Here $F=f$, which is not outer semicontinuous.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.