- #1
kaosAD
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Hi,
Let [tex]C_1[/tex] and [tex]C_2[/tex] be nonempty convex sets and suppose [tex]C_1 \cap C_2 \neq \emptyset [/tex]. I read a text that claims [tex]\textup{cl}(C_1 \cap C_2) \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex] since [tex]C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex].
I am able to prove the latter, but I am not able to see the deduction made by the author. Hope someone can help me out.
Let [tex]x \in C_1 \cap C_2 \neq \emptyset [/tex], then [tex]x \in C_1 [/tex] and [tex] x \in C_2[/tex]. This implies [tex]x \in \textup{cl}(C_1) [/tex] and [tex]x \in \textup{cl}(C_2)[/tex]. Hence [tex]x \in \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex]. So [tex]C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex].
p/s: cl means closure
Let [tex]C_1[/tex] and [tex]C_2[/tex] be nonempty convex sets and suppose [tex]C_1 \cap C_2 \neq \emptyset [/tex]. I read a text that claims [tex]\textup{cl}(C_1 \cap C_2) \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex] since [tex]C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex].
I am able to prove the latter, but I am not able to see the deduction made by the author. Hope someone can help me out.
Let [tex]x \in C_1 \cap C_2 \neq \emptyset [/tex], then [tex]x \in C_1 [/tex] and [tex] x \in C_2[/tex]. This implies [tex]x \in \textup{cl}(C_1) [/tex] and [tex]x \in \textup{cl}(C_2)[/tex]. Hence [tex]x \in \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex]. So [tex]C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)[/tex].
p/s: cl means closure
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