I was wondering if my proof/solution for i & ii was fine. Any feedback is greatly appreciated, please. Thank you. $\def\R{{\mathbb R}} % real numbers \def\N{{\mathbb N}}$

Let $E_1\supseteq E_2\supseteq \,\cdots\,\supseteq E_k\supseteq \,\cdots$ be a decreasing sequence of nonempty, closed subsets of $\R^n$.

(i) Prove if $E_1$ is compact, then $\displaystyle{\bigcap_{n=1}^\infty E_n}\ne\emptyset$.

$\textbf{Proof:}$ Assume $E_1$ is compact. Since each $E_n$ is non-empty, choose a point $x_n \in E_n$ for all $n\in\N$. Then $\{x_n\}$ be a sequence in $\R^n$. Now, each $E_n\subseteq E_n$ for all $n\ge 2$. So, $x_n \in E_1$ for all $n\in \N$.

Now, $E_1$ is compact. So $\{x_n\}$ has a convergent subsequence. Let $\{x_{n_k}\}$ be the subsequence of $\{x_n\}$ which converges to $x\in \R^n$.

Our claim is that $\displaystyle{x\in \bigcap_{n=1}^\infty E_n}$. As $x_{n_k} \to x$, so every neighbourhood of $x$ contains all, but excepting finitely many possibly.

Now, for all $n\in \N, x_m\in E_n$ for all $m\ge n$ as $x_m \in E_m \subseteq E_n$ for all $m\ge n$ implies $x_{n_k} \in E_n$ for all $m\ge n$. Therefore, as $n_m \ge m$ implies each neighbourhood of $x$ intersects $E_n$ and implies $x\in \overline{E_n} = E_n$, $E_n$ is closed, for all $n\in \N$. So, $\displaystyle{x\in \bigcap_{n=1}^\infty E_n}$.

(ii) Provide an example of a sequence as above with the property $\displaystyle{\bigcap_{n=1}^\infty E_n}=\emptyset$.

$\textbf{Solution:}$ Take $E_k = \N \setminus \{1,2,3,\dots , k\}$ for all $k\in \N$. Then each $E_k$ is a closed subset of $\R$. Also, $E_1\supseteq E_2\supseteq \dots$ . Now, $$\bigcap_{n=1}^\infty E_n = \N\setminus \{1,2,3, \dots\} = \emptyset.$$ Thus, $\{E_n\}$ is a sequence of closed nonempty subsets of $\R$ with $\displaystyle{\bigcap_{n=1}^\infty E_n = \emptyset}.$

  • $\begingroup$ I think it's easier to note that all $E_i$ are compact since closed subsets of compact sets are compact, and it's quite well known that a decreasing series of compact sets has non-empty intersection. $\endgroup$ Commented Apr 4, 2020 at 15:24
  • $\begingroup$ Your proof is correct. If your only available definition of compactness is that every sequence has a convergent subsequence, it’s a perfectly reasonable choice. Note, though, that that is actually the definition of a different property, sequential compactness, that is not equivalent to the general topological notion of compactness, though the two notions are equivalent in $\Bbb R^n$. There is a much easier argument using the usual general definition of compactness (every open cover of the space has a finite subcover). $\endgroup$ Commented Apr 4, 2020 at 15:31
  • $\begingroup$ A simple example: $E_k=[k+\infty)$ with $k\in\mathbb N$. $\endgroup$
    – John B
    Commented Apr 6, 2020 at 13:10
  • $\begingroup$ If the main point of your post is to get feedback on your proof, you can use the (solution-verification) tag to indicate this. $\endgroup$ Commented Apr 28, 2020 at 9:19

3 Answers 3


Your proof is correct, if your definition of compact is sequential compactness (i.e., in terms of sequences).

What you have proved here is known as Cantor's intersection theorem, you can compare your proof with that on Wikipedia which is basically the same.


Choosing $ x_{n}\in E_{n},\ \left(\forall n\in\mathbb{N}\right)\cdot $ And $ \left(x_{\varphi\left(n\right)}\right) $ the subsequence of $ \left(x_{n}\right) $ that converges to $ x\in E_{1} \cdot $

If $ n\in\mathbb{N} $, then since $ \left(\forall p\geq n\right),\ E_{p}\subset E_{n} $, we have that $ \left(x_{p}\right)_{p\geq n} $ is a sequence of points of $ E_{n} \cdot $ And since it does converge to $ x $ and $ E_{n} $ is closed, then $ x\in E_{n} \cdot $

Since $ n $ is arbitrary in $ \mathbb{N} $, then $ \left(\forall n\in\mathbb{N}\right),\ x\in E_{n} $, thus $ x\in\bigcap\limits_{n\in\mathbb{N}}{E_{n}} $, meaninig $$ \bigcap_{n\in\mathbb{N}}{E_{n}}\neq\emptyset $$


This works in every Hausdorff space, where compact implies closed.

It is not restrictive to assume that the whole space is $E_1$. Define $A_i=E_1\setminus E_i$, so $A_i$ is open. If $$ \bigcap_{i\ge1}E_i=\emptyset $$ then also $$ \bigcup_{i\ge1}A_i=E_1 $$ Since $A_i$ is open in $E_i$ which is compact, there is a finite subcover. As the sequence $A_i$ is increasing, this implies that $A_k=E_1$ for some $k$, hence $E_k=\emptyset$.

You can simplify your proof by observing that, once you have chosen the subsequence $x_{n_k}\to x$, you have $$ \bigcap_{k}E_{n_k}=\bigcap_{n}E_n $$ Hence it is not restrictive to assume that $x_n\to x$ to begin with.

If $x\notin E_n$, for some $n$, then there exists a neighborhood $U$ of $x$ such that $E_n\cap U=\emptyset$. But then $x_m\notin U$, for every $m>n$, contradicting convergence.


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