Let $A\subset \mathbb{R}$ be a compact set.
Define $$B_j=\left\{x\in\mathbb{R} : \text{dist}(x, A)<\frac{1}{j}, j\in\mathbb{N}\right\}$$ Prove that $$\bigcap_{j=1}^\infty B_j\subset A$$ Proof:
Suppose $$\bigcap_{j=1}^\infty B_j\not\subset A\\\implies \exists x\in\bigcap_{j=1}^\infty B_j\setminus A\\\implies \text{dist}(x,A)>\epsilon\enspace\text{ for some }\epsilon >0$$ By Archimedean property $\exists j_0\in\mathbb{N}$ such that $\epsilon>\frac{1}{j_0}$.$$\implies \text{dist}(x,A)>\frac{1}{j_0}\\\implies x\not\in B_{j_0}$$This is a contradiction to the fact that $x\in\bigcap_{j=1}^\infty B_j$.
Hence, $\bigcap_{j=1}^\infty B_j\subset A$.

Is my proof correct? If not, please provide me the mistake and how to fix it?
Also, If there are any alternative proof?

  • 3
    $\begingroup$ Yes, this is correct. Note that it actually shows that $\bigcap_{j\ge 1}B_j=A$, since the intersection clearly contains $A$. $\endgroup$ Jun 19, 2020 at 22:29
  • $\begingroup$ @BrianM.Scott I am unable to comprehend the use of compactness of A given in the question. Can you please point it out? $\endgroup$
    – Kumar
    Jun 19, 2020 at 22:34
  • 1
    $\begingroup$ You don't need compactness. $\endgroup$ Jun 19, 2020 at 22:36
  • 1
    $\begingroup$ You don’t actually need $A$ to be compact: you just need it to be closed, so that the distance from $x$ to $A$ is positive. Thus, compactness is used here only to ensure that $A$ is closed. $\endgroup$ Jun 19, 2020 at 22:37
  • $\begingroup$ Maybe I’m missing something, but how did you come to the implication: $\exists x in \cap_{j\geq 1}B_j\backslash A \implies dist(x,A)>\epsilon$, for some $\epsilon>0$. I’m confused since $x\in B_j$ for all $j$, and thus $dist(x,A)<\frac{1}{j}$ for all $j$. Taking limits we arrive that $lim_{j\rightarrow 0} d(x,A) = 0$, which implies that $x\in A$. Hence a contradiction. $\endgroup$
    – user798642
    Jun 20, 2020 at 5:37

1 Answer 1


Direct proof

Let $ x\in \Bbb R$.

$$x\in \bigcap_{n\ge 1}B_n \implies$$

$$(\forall n\ge 1)\; \;\; x\in B_n \implies$$

$$(\forall n\ge 1) \;\;\; 0\le d(x,A)<\frac 1n \implies$$

$$\lim_{n\to+\infty}d(x,A)=0 \implies$$ $$d(x,A)=0 \implies x\in A$$ because $ A $ is closed as pointed by @Brian M. Scott,

thus $$\bigcap_{n\ge 1}B_n\subset A$$


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