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Let $M$ be a subspace of a metrizable space $X$ and $U_i$, $i\in I$, are open sets in $M$. Prove that there exist open sets $V_i$, $i\in I$, in $X$ such that $U_i=V_i\cap M$ and for any finite $F\subseteq I$, if $\bigcap_{i\in F}U_i=\emptyset$ then $\bigcap_{i\in F}V_i=\emptyset$.

I am clueless about this problem.

Edit: No correct answer so far (two incorrect proofs were deleted). I think the idea is to write $V_i=\bigcup_{x\in U_i}B(x,r_{ix})$ for sufficiently small $r_{ix}>0$. But the difficulty lies in choosing $r_{ix}$ to capture the desired conclusion that $\bigcap_{i\in F}U_i=\emptyset\implies\bigcap_{i\in F}V_i=\emptyset$ for any finite $F\subseteq I$.

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  • $\begingroup$ What sort of space is $X$? Note that if $X = \{ 0 , 1 , 2 \}$ is given the topology $\{ \varnothing , \{ 0 , 2 \} , \{ 1 , 2 \} , \{ 2 \} , X \}$ then the subspace $M = \{ 0 , 1 \}$ does not have the required properties. ($\{ 0 \} = \{ 0 , 2 \} \cap M$ and $\{ 1 \} = \{ 1 , 2 \} \cap M$ are disjoint open sets in $M$, but $\{ 0 , 2 \} \cap \{ 1 , 2 \} \neq \varnothing$.) $\endgroup$ – user642796 Mar 12 '13 at 7:01
  • $\begingroup$ I forgot to say that $X$ is metrizable $\endgroup$ – abc Mar 12 '13 at 9:08
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For each $i \in I$ define $$V_i = \left\{ x \in X : d ( x , U_i ) < \tfrac{1}{2} d ( x , M \setminus U_i ) \right\}.$$ The sets $V_i$ are clearly open in $X$, and $V_i \cap M = U_i$.

Let $F \subseteq I$ be finite such that $\bigcap_{i \in F} U_i = \emptyset$. Suppose that $x \in \bigcap_{i \in F} V_i$. We can recursively construct sequences $\langle i_k \rangle_{k=0}^\infty$ and $\langle y_k \rangle_{k=0}^\infty$ so that

  • $i_k \in F$;
  • $y_k \in U_{i_k}$;
  • $d ( x , y_k ) < \frac{1}{2} d ( x , M \setminus U_{i_k} )$; and
  • $y_k \notin U_{i_{k+1}}$

As $d ( x , y_{k+1} ) < \frac{1}{2} d ( x , M \setminus U_{i_{k+1}} ) \leq \frac{1}{2} d ( x , y_k )$ it is easy to see that $\lim_{k \rightarrow \infty} d ( x , y_k ) = 0$. As $F$ is finite there is an $i \in F$ and a strictly increasing sequence $\langle k_j \rangle_{j=0}^\infty$ such that $i_{{k_j}+1} = i$ for all $j$. As $y_{k_j} \in M \setminus U_{{k_j}+1} = M \setminus U_i$ for all $j$ it follows that $d ( x , M \setminus U_i ) \leq \lim_{j \rightarrow \infty} d ( x , y_{k_j} ) = 0$, contradicting that $x \in V_i$!

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  • $\begingroup$ Thanks a lot! Just a small typo after the bullets where it should be $\frac12d(x,M\setminus U_{i_{k+1}})$. Just out of curiosity, how did you come up with this solution? $\endgroup$ – abc Mar 14 '13 at 5:59
  • $\begingroup$ @abc: I'll correct the typos in a bit. My original attempt was looking at $d(x,U)<d(x,M\setminus U)$ (which was "something Brian Scott didn't try in his deleted attempt" -- after 10K reputation you can see deleted posts). The construction of sequences came about because... I could? I was just playing around, really, and couldn't think of another avenue of approach. And then I realised that I got a nice contradiction if the $y_i$ converged to $x$, so I altered the definition of the open sets to ensure that this happened. So basically I fiddled around and got lucky. That's math! :-) $\endgroup$ – user642796 Mar 14 '13 at 6:22

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