The set of Upper semi-continuous functions as a ring. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:03:27Z http://mathoverflow.net/feeds/question/106190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106190/the-set-of-upper-semi-continuous-functions-as-a-ring The set of Upper semi-continuous functions as a ring. AliReza Olfati 2012-09-02T17:24:35Z 2012-09-02T17:41:40Z <p>I should recall that the <strong>surgenfery topology</strong> on the real numbers is denoted by $\mathbb{R}_l$, and has the set<br> {$[a , b): a,b \in \mathbb{R}$} as it's base.</p> <p>If $X$ is a topological space, an upper semi-continuous real function on $X$ can be interpret as a <strong>continuous function</strong> from $X$ into $\mathbb{R}_l$. </p> <p>The set of all <strong>upper semi-continuous real functions</strong> on $X$ is denoted by $USC(X)$.</p> <p>We could easily see that if we consider $X=\mathbb{R}$ then the set $USC(\mathbb{R})$ is not a ring with pointwise addition and multiplication. because there is $f \in USC(\mathbb{R})$ so that $-f \notin USC(\mathbb{R})$. <strong>Indeed it is not a Group</strong>.</p> <p><strong>Question</strong>.For what condition(s) on $X$, the set $USC(X)$ constructs a ring structure we the pointwise addition and multiplication?</p> <p>PS:I am looking for the topological condition(s) $P$ on $X$ so that, $USC(X)$ is a ring iff $X$ has the property $P$.</p> <p>Thank you so much for noticing to my Question. </p> http://mathoverflow.net/questions/106190/the-set-of-upper-semi-continuous-functions-as-a-ring/106192#106192 Answer by Will Sawin for The set of Upper semi-continuous functions as a ring. Will Sawin 2012-09-02T17:41:40Z 2012-09-02T17:41:40Z <p>Any set $U$ which is open and not closed provides an example: the characteristic function $\chi_U$ is in $USC(X)$, but $-\chi_U$ is not, or perhaps the other way around.</p> <p>Topologies where every open set is closed, and vice versa, come from partitions of the base set, with open and closed sets being arbitrary unions of parts. Thus, a semicontinuous function from such a topological space to $\mathbb R$ is just a function $X \to \mathbb R$ that is constant on the parts, or a function from the set of parts to $\mathbb R$.</p> <p>Since the space of functions from any set to $\mathbb R$ form a ring under pointwise addition and multiplication, $USC(X)$ is a ring if and only if the topology on $X$ is of this type.</p>