If $E\subseteq X$ is connected, $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$

Question

Let $X$ be topological space and $A\subseteq X$.

If $E\subseteq X$ is connected, $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$.

Assume that $E\cap\partial A=\emptyset$.

If $E\cap\partial A=\varnothing$ then $E=(E\cap\text{int}(A))\cup(E\cap\text{ext}(A))$, i.e. a union of disjoint sets open in $E$.

Also the sets are nonempty because: $E\cap\text{int}(A)=E\cap A\neq\varnothing$ and $E\cap\text{ext}(A)=E\cap A^c\neq\varnothing$.

This contradicts that $E$ is connected.

Fixed.

I don't understand what you wanted to say in the sentence "Now, because $E\cap A\neq\emptyset$ and $E\cap(X\setminus A)\neq\emptyset$ then $E\cap\partial A\neq\emptyset$, we have $E\subset int(A)$ and $E\subseteq ext(A)$. " — 5xum, Mar 10 '16 at 9:14

drhab · Accepted Answer · 2016-03-10 09:34:09Z

1

If $E\cap\partial A=\varnothing$ then $E=(E\cap\text{int}(A))\cup(E\cap\text{ext}(A))$, i.e. a union of disjoint sets open in $E$.

Also the sets are nonempty because: $E\cap\text{int}(A)=E\cap A\neq\varnothing$ and $E\cap\text{ext}(A)=E\cap A^c\neq\varnothing$.

This contradicts that $E$ is connected.

answered Mar 10 '16 at 9:23

drhab

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