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Consider the real line $\mathbb{R}$. Is there an open set in $\mathbb{R}$ that is not almost closed? I am using the term "almost closed" to mean a closed set modulo a finite set, i.e, a closed set plus and/or minus a finite set of points.

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Consider $U=(0,1) \backslash \{1/n,\, n \geq 1\}$. Assume there is a closed subset $S \subset \Bbb R$ such that $U \Delta S$ is finite. This means that (up to adding finitely many points to $S$) there is a finite set $F$ of real numbers such that $U \cup F =S$. In particular, $S \supset \overline{U}=[0,1]$, so $F \supset [0,1] \backslash U$ which is infinite – a contradiction.

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Consider $$ \mathbb{R}\setminus \mathbb{Z} $$

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    $\begingroup$ What is the proof that it is not almost closed? I already know it is open. $\endgroup$ – user107952 Jan 27 at 22:59
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    $\begingroup$ @user107952 $\mathbb{Z}$ is not finite; the closure of this set is $\mathbb{R}$. $\endgroup$ – user9464 Jan 27 at 23:01
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Since some of the other answers are making assertions without proof, let's actually flesh these arguments out. We say $A$ is almost closed if there exists a closed set $F$ such that $F\triangle A$ is finite. On the reals, which are a $T_1$ space, finite sets are closed, so $F':=F\cup(F\triangle A)$ is also a closed set; thus we can restrict attention to $F\supseteq A$ such that $F\setminus A$ is finite.

Now if $A\subseteq F$, then $\bar A\subseteq F$, so $\bar A$ is also a closed set that differs from $A$ by finitely many points. So an equivalent characterization of almost closed sets $A$ is that $\bar A\setminus A$ is finite.

So now it is easy to see that $\Bbb R\setminus\Bbb Z$ is not almost closed; it is open (because it is equal to $\bigcup_{i\in\Bbb Z}(i,i+1)$), but the closure of $\Bbb R\setminus\Bbb Z$ is $\Bbb R$ because every point in $\Bbb Z$ is a limit point of $\Bbb R\setminus\Bbb Z$ (to wit, $i+1/n$ for $n\in \Bbb N^{>1}$ is a sequence in $\Bbb R\setminus\Bbb Z$ that converges to $i\in\Bbb Z$).

Now let us consider extending the cardinality in the definition of almost closed. Is there an open set $A$ such that if $F$ is any closed set then $F\triangle A$ is cardinality $\frak c$? That's certainly the most we can hope for on $\Bbb R$, and @TippingOctopus suggests that an example is the complement of the Cantor set.

Since the Cantor set $C$ is closed and uncountable, its complement $A:=\Bbb R\setminus C$ is open, and $\bar A=\Bbb R$ because $C$ has empty interior. However, the earlier reduction to closures doesn't immediately apply here because while finite sets are closed, sets of larger cardinality may not be, so we can't just assume that the only set $F$ of interest is $\bar A$. If we take a countable (or even cardinality $<\frak c$) set of points out of $C$, can we construct something with a different closure?

It turns out that the answer is no. Suppose $S$ is any set of cardinality $<\frak c$. Because $A$ is dense and open (although you need far less than this), for any open neighborhood $V$ of a point $x\in A$, $V\cap A$ is nonempty (because $A$ is dense) and open, so it contains an interval and hence is cardinality $\frak c$; thus $V\cap A\setminus S$ is nonempty. Therefore, $x$ is a limit point of $A\setminus S$, and hence $\overline{A\setminus S}\supseteq A$ (so in fact $\overline{A\setminus S}=\Bbb R$).

To apply this theorem to prove that $A$ is not-$\frak c$-almost closed, let $F$ be a closed set, let $S=F\triangle A$ and suppose $|S|<\frak c$. Then as just shown $A\setminus S=A\cap F$ is dense, so $F$ is also dense and hence $F=\bar F=\Bbb R$. Thus $|S|=|C|=\frak c$ after all, a contradiction.

Thus the complement of the Cantor set is an open set of reals that differs from any closed set by a set of cardinality $\frak c$.

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The difference between an open subset of $\mathbb{R}$ and its closure can be huge. To built an example, let us start with an enumeration $(q_n)_{n \in \mathbb{N}}$ of all rational numbers. Let $\varepsilon > 0$ be some tiny positive number. Now consider the following open set:

$$U := \bigcup_{n \in \mathbb{N}} (q_n - \varepsilon 2^{-n-2}, q_n + \varepsilon 2^{-n-2})$$

As $\mathbb{Q} \subseteq U$, we know that $U$ is dense, so $\mathrm{cl} U = \mathbb{R}$. On the other hand, we also know that $\lambda(U) \leq \varepsilon$. It follows that $\mathrm{cl} U \setminus U$ has unbounded Lebesgue measure (so in particular, is uncountable). We can also make it have arbitrarily small positive density in $[0,1]$.

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The other answers have shown that there are open sets which do not differ from a closed set by any set of cardinality less than $2^{\aleph_0}$. I'm surprised that no one has yet mentioned the "correct" way to fix your statement, which is to replace cardinality by Baire category as our notion of "smallness".

Let $U$ be an open set. Then $F = \mathrm{cl}(U)\setminus U$ has empty interior. [Proof: Suppose $V\subseteq F$ is open. Then $U\subseteq \text{cl}(U)\setminus V \subseteq \text{cl}(U)$ and $\text{cl}(U)\setminus V$ is closed, so $\text{cl}(U)\setminus V = \text{cl}(U)$, hence $V$ is empty.] Since $F$ is closed and has empty interior, $F$ is nowhere dense (not dense in any open set). So every open set is "almost closed", where a set is "almost closed" if it is a closed set modulo a nowhere dense set.

From the point of view of Baire category, the nowhere dense sets are the smallest sets. Enlarging this category a bit, we call a set meager if it is a countable union of nowhere dense sets. It turns out that the equivalence relation "equal modulo a meager set" is a very robust and useful one. For the rest of this answer, I'll interpret "almost" to mean "modulo a meager set".

A set has the Baire property if it is almost open. It turns out that the collection of sets with the Baire property is closed under complements, and it follows that every set with the Baire property is also almost closed. Further, the collection of sets with the Baire property is closed under countable unions, which makes it a $\sigma$-algebra. So many many sets of reals are almost closed, including every Borel set.

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