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I got the following question and solution from somewhere else. Now, I understand why the set $[0,1)$ is clopen but I don't understand the solutions reasoning. Is there some theorem that says the intersection of an open (closed) set with any set is always open (closed)?

Question:

Consider $X=[0,2]\setminus{1}$ as a subspace of the real line $\mathbb{R}$. Show that the subset $[0,1)\subset X$ is both open and closed.

Answer:

$[0,1)=(-\infty,1)\cap X$. Since $(-\infty,1)$ is open $[0,1)$ is open in $\mathbb{R}$.

$[0,1)=[0,1]\cap X$. Since $[0,1]$ is closed $[0,1)$ is closed in $\mathbb{R}$.

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    $\begingroup$ There is the definition of subspace topology on $X\subset M$ that states that a set is open in $X$ iff it is the intersection of $X$ and an open set in $M$. $\endgroup$ – joedoe8585 Jan 22 '16 at 20:22
  • $\begingroup$ Yeah, that's what I get for reading top-down. :-) $\endgroup$ – Brian Tung Jan 22 '16 at 20:24
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There is no such theorem, because it's false. What's going on in the solution is the definition of the subspace topology:

Given a topological space $(A, \tau)$ and a subset $X\subseteq A$, the subspace topology on $X$ is the topology given by open sets in $A$ intersected with $X$: $(X, \{U\cap X: U\in\tau\})$.

That is, a set $Y\subseteq X$ is open in the sense of $X$ if $Y=X\cap U$ for some $U$ which is open in the sense of $A$. Similarly, we can prove that a set $Y\subseteq X$ is closed in the sense of $X$ if $Y=X\cap C$ for some $C$ which is closed in the sense of $A$.

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