Removing finitely many point from an open set in $\mathbb{R}^n$ gives an open set. Is this true in general for any space?

My intuition is that this is the case, however, how does one (dis)prove this?

The only idea that comes in mind is that, since singletons are closed and unions of closed sets are closed, a union of singletons is closed.

Now, let $U$ be an open set. Then the complement of $U$ is closed, and any point $x$ removed from $U$ is a singleton that is "unioned" with the closed complement of $U$ to form a bigger closed set $C$. $C$ is the complement of $U\setminus\{x\},$ so $U\setminus\{x\}$ is open.

One can thus repeat this argument inductively for any finite number of removed points.

Does this make sense?

  • 1
    $\begingroup$ Just a small note that if $U\subseteq X$ is open and $C\subseteq C$ is closed, then $U\setminus C = U\cap(X\setminus C)$ is the intersection of two open sets and is therefore open. Your observation in $\mathbb{R}^n$ is a special case of this. $\endgroup$ Nov 17, 2019 at 17:30

3 Answers 3


The argument makes sense in general topological spaces where singletons are closed ($T_1$ spaces I think).

If some singleton $\{x\}$ is not closed in the space $X$, then $X$ is open in $X$ while $X \backslash \{x\}$ is not.

  • 4
    $\begingroup$ +1 this is a good remark $\endgroup$ Nov 16, 2019 at 23:02
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    $\begingroup$ In fact, OPs property is equivalent to being $T_1$. $\endgroup$
    – Wojowu
    Nov 17, 2019 at 9:55

If you allow arbitrary topologies ("any general space"), then the answer is no. For any set $M$ with at least two elements you can define a topology $\mathcal T=\{\emptyset, M\}$ (i.e. only the empty set and $M$ are open sets in $M$).

Then, by removing one point from $M$, you get a set which is not open.

Edit: Let me add the following:

Proposition. Let $(M,\mathcal T)$ be a topological space. Then the following two statements are equivalent:

  1. For every $x\in M$, $\{x\}$ is closed (i.e. $M\setminus\{x\}\in\mathcal T$),
  2. For every open set $O\in \mathcal T$ and point $x\in M$, we have that $O\setminus\{x\}$ is also open.

Proof. 1.$\implies$2.: Suppose that 1. is true and let $O\in \mathcal T$, $x\in M$. Then $M\setminus\{x\}$ is open and hence $$O\setminus\{x\}=O\cap\big(M\setminus\{x\}\big)\in\mathcal T.$$

2.$\implies$ 1.: Suppose that 2. is true and let $x\in M$. Then, since $M$ is open, $M\setminus\{x\}$ is open too, so $\{x\}$ is closed. $\square$

  • $\begingroup$ Indeed. Is there a nontrivial counterexample to this? $\endgroup$
    – Stephen
    Nov 16, 2019 at 22:58
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    $\begingroup$ @Stephen If you want, you can consider $M=\{a,b,c,d\}$ and $\mathcal T=\{\emptyset,\{a,b\},\{c,d\},M\}$. The intuition is to avoid singletons being closed sets $\endgroup$ Nov 16, 2019 at 23:03
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    $\begingroup$ @stephen The Sierpiński space is the smallest nontrivial example. Its open sets are $\{a,b\}, \{a\}, \varnothing$. $\endgroup$
    – MJD
    Nov 17, 2019 at 15:18
  • $\begingroup$ Nice addition. Though I feel silly now for not immediately seeing this before asking my question. I guess I was so focused on my intuitions that I didn't even consider looking for a counterexample! $\endgroup$
    – Stephen
    Nov 17, 2019 at 20:10

The reason this works in $\Bbb R^n$ is that a set consisting of a single point is always closed. Then you can understand $$G\setminus\{ x_1, x_2, \ldots, x_k \}$$ as $$G\cap (\Bbb R^n \setminus \{x_1\}) \cap (\Bbb R^n \setminus \{x_2\})\cap\ldots \cap (\Bbb R^n \setminus \{x_k\}).$$ This is a finite intersection of open sets, so is open.

In general spaces, a singleton set $\{x_1\}$ may not always be closed. A space in which every singleton is closed is called a $T_1$ space. All metric spaces are $T_1$, but it is easy to construct quite simple spaces that are not $T_1$. For example, let $S=\{a,b,c\}$ and let the open sets be $S, \{a,b\}, \{a,c\}, \{a\}, \varnothing$. Then $\{c\}$ is closed but $\{a\}$ is not, and indeed if you remove the single point $a$ from the open set $\{a,b\}$ you get $\{b\}$, which is not an open set.


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