When an intersection of open sets is a singleton

I need to prove that if F={x}⊂RnF=\{x\}\subset \mathbb{R}^n then ∃\exists a collection of open subsets UiU_i, i.e. {U1,U2,…,…}\{U_1, U_2,…,…\}, such that ⋂iUi={x}\bigcap_i U_i = \{x\}.

My approach:

Consider the collection of open balls B1/n(x)B_{1/n}(x), for all n∈Nn\in\mathbb{N}. Then ∞⋂n=1B1/n(x)={x}\bigcap\limits_{n=1}^\infty B_{1/n}{(x)}=\{x\}.

Do you think this is rigorous enough, or should there be further clarification in the proof?

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Looks good to me.
– Antioquia3943
Oct 21 at 1:51

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Your solution is correct but your teacher may require a proof that your solution is correct. You must show that xx is in the intersection and that if y≠xy\ne x then it is not in the intersection.
– John Douma
Oct 21 at 2:37

  

 

@JohnDouma, would it be OK to argue like this? Let y∈∞⋂1/nB1/n(x)y\in \bigcap\limits_{1/n}^\infty B_{1/n}(x). Then y∈limn→∞B1/n(x)=B0(x)={x}y\in \lim\limits_{n\to\infty} B_{1/n}(x)=B_0(x)=\{x\}. I’m a bit concerned about the B_0(x)B_0(x).
– sequence
Oct 21 at 2:47

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x\in B_{\frac{1}{n}}(x)x\in B_{\frac{1}{n}}(x) for all nn so xx is in the intersection. If y\ne xy\ne x then choose NN such that \frac{1}{N}\lt d(x,y)\frac{1}{N}\lt d(x,y). Then y\notin B_{\frac{1}{N}}(x)y\notin B_{\frac{1}{N}}(x) so yy is not in the intersection.
– John Douma
Oct 21 at 3:24

  

 

It depends who your audience is. Note that the result is valid in any metric space.
– user254665
2 days ago

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