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$T$ is said to have local intersection property if for each $x\in X$ with $T(x)\neq\emptyset$, there exists an open neighborhood $N_{x}$ of $x$ such that $\cap_{z\in N_{x}}T(z)\neq\emptyset$.

Consider $E=F=[0,2)$ and define a correspondence $T\colon E\twoheadrightarrow F$ by $T(x)=[x,2)$.

How can I show that it has a local intersection property?

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  • $\begingroup$ It would be extremely helpful, say for your question, if you'd say what in the world is that $\;T\;$ there... $\endgroup$ – DonAntonio Mar 6 '18 at 10:41
  • $\begingroup$ @DonAntonio Sorry i forget to mention, $T$ is a correspondence from $X$ to $Y$ where $X$ and $Y$ are linear topological spaces. $\endgroup$ – User70020 Mar 7 '18 at 11:17
  • $\begingroup$ Good...but then what does $\;T(x)\neq\emptyset\;$ mean, anyway? The correspondence seems to be between sets...And how would you make $\;[0,2)\;$ (which I presume is the half open interval on the real line...) a topological linear space, or part of one? I think you need to add lots of more info, definitions, clarifications, etc. to this question... $\endgroup$ – DonAntonio Mar 7 '18 at 11:39
  • $\begingroup$ @DonAntonio $T(x)$ is a non-empty subset of $Y$. $T\colon[0,2)\rightarrow{2^{[0,2)}}$ define by $T(x)=[x,2)$. $\endgroup$ – User70020 Mar 7 '18 at 11:45
  • $\begingroup$ But then $\;T\;$ is not a correspondence between two linear top. spaces, but a map from the set of subsets of one to the set of subsets of the other one...? A usual map defined from a set $\;A\;$ to a set $\;B\;$ has elements of $\;B\;$ as images...how come you get $\;Tx=\emptyset\;$ ?? $\endgroup$ – DonAntonio Mar 7 '18 at 13:40

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