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Prove that in every metric space $X$ , the set $X$ \ {x} is an open set in X.

I started by suposing the opposite that $X$ \ {x} is a closed set in $ X $ which means tht the complement $ ( X $ \ {x})$ ^c$ = {x} is an open set , which is a contradiction. Im not sure if i have the right contradiction . Can anyone help me ? Thank you !

PseudoNeo
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MATH14
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    The opposite of "$X \setminus {x}$ is open" is not "$X \setminus {x}$ is closed". A set can be neither open nor closed. For example, the half-open interval $(a,b] \subseteq \mathbb{R}$ is neither open nor closed in the usual topology on $\mathbb{R}$. –  Oct 14 '15 at 18:02
  • @Bungo is right. Let's imagine for example the set X={a,b} with the topology: ${{a,b},a,b, \emptyset }$ it is clear that b and X{b} are both open and both closed. And they are "opposite".

    Hint for solving your problem: all metric spaces are Hausdorff

     – D1811994 Oct 14 '15 at 18:09
  • Alternatively, if you have available the standard fact that the distance function $d: X \times X \to [0, \infty)$ is continuous, so its restriction to ${x} \times X$, but $X \setminus {x}$ is just the preimage of the open set $(0, \infty)$ under this restriction. – Travis Willse Oct 14 '15 at 18:33

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It’s entirely possible for $\{x\}$ to be an open set. For example, take $X$ to be $\Bbb Z$, the set of integers, with the usual metric. Then for each $n\in\Bbb Z$ we have $\{n\}=\{m\in\Bbb Z:|m-n|<1\}$, the open ball of radius $1$ centred at $n$, which is certainly an open set.

You should instead try to prove directly that $X\setminus\{x\}$ is open. To do this, let $y\in X\setminus\{x\}$, and try to find an $\epsilon>0$ such that $B(y,\epsilon)\subseteq X\setminus\{x\}$. Note that this is the same as saying that $x\notin B(y,\epsilon)$; can you find an $\epsilon$ for which this is true? Note that the choice of $\epsilon$ will depend on $y$.

Brian M. Scott
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The opposite of "$S$ is open" is not "$S$ is closed". There are sets that are neither closed nor open, and sets that are both.

The real point to consider here is that the set $\{x\}$ is closed. This will be true in every metric space. Sometimes, it will also be open.

Ben Grossmann
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You are wrong (twice).

Firstly, the "opposite" of open is not closed.

Secondly, $\{x\}$ can be open. Consider the metric space $X=\{x\}$.

What you want to do is prove that every point-set is closed. This follows easily from hausdorffness, which is a property of metric spaces. Since point-sets are closed, the complement of closed is open, finite intersection of open sets is open and the whole metric space is open, you have that $X\backslash \{x\}=X \cap \{x\}^c$ is open.

Aloizio Macedo
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