[proofplan] We replace each [nowhere dense set](/page/Nowhere%20Dense%20Set) by its closure, so the problem is to find one real number avoiding a family of closed sets with empty interior. We force with rational open intervals ordered by reverse inclusion: meeting the dense set attached to $i \in I$ makes some interval avoid $\overline{A_i}$, while meeting the length dense sets makes the intervals arbitrarily small. The filter is directed rather than linearly ordered, so we use directedness to show that the selected closed intervals have a unique common point, and the avoidance dense sets then put this point outside every $A_i$. [/proofplan] [step:Replace the nowhere dense sets by closed nowhere dense sets] For each $i \in I$, define $F_i := \overline{A_i}$, where the closure is taken in the usual topology on $\mathbb{R}$. Since $A_i$ is nowhere dense, $F_i$ has empty interior. Since $A_i \subset F_i$, it is enough to prove \begin{align*} \mathbb{R} \neq \bigcup_{i \in I} F_i. \end{align*} Indeed, any point of $\mathbb{R} \setminus \bigcup_{i \in I} F_i$ also belongs to $\mathbb{R} \setminus \bigcup_{i \in I} A_i$. [guided] The family in the statement consists of nowhere dense sets $A_i$, which need not be closed. It is more convenient to work with closed sets because later we will choose an interval whose closure avoids one of them. For each $i \in I$, define $F_i := \overline{A_i}$, with closure taken in the usual topology on $\mathbb{R}$. The definition of nowhere dense says exactly that $\overline{A_i}$ has empty interior, so each $F_i$ is closed and has empty interior. Also $A_i \subset F_i$ for every $i \in I$. Therefore, if we can find a point outside every $F_i$, then that same point is outside every $A_i$. In symbols, it is enough to prove \begin{align*} \mathbb{R} \neq \bigcup_{i \in I} F_i. \end{align*} Indeed, any $x \in \mathbb{R} \setminus \bigcup_{i \in I} F_i$ satisfies $x \notin F_i$ for every $i \in I$, and since $A_i \subset F_i$, it follows that $x \notin A_i$ for every $i \in I$. [/guided] [/step] [step:Define the interval forcing and verify the ccc condition] Let $\mathbb{Q}$ denote the set of rational numbers. Let $P$ be the set of all non-empty bounded open intervals with rational endpoints: \begin{align*} P := \{(a,b) \subset \mathbb{R} : a,b \in \mathbb{Q} \text{ and } a < b\}. \end{align*} Define the partial order $\leq$ on $P$ by reverse inclusion: \begin{align*} p \leq q \quad \Longleftrightarrow \quad p \subset q. \end{align*} Thus a stronger condition is a smaller interval. The set $P$ is countable because $\mathbb{Q}^2$ is countable and $P \subset \mathbb{Q}^2$ under the endpoint parametrisation $(a,b) \mapsto ((a,b))$. Therefore every antichain in $P$ is countable, so $P$ satisfies the countable chain condition. [guided] We need a poset satisfying the countable chain condition in order to apply $MA_\kappa$. Let $\mathbb{Q}$ denote the set of rational numbers, and define $P$ to be the set of all non-empty bounded open intervals with rational endpoints: \begin{align*} P := \{(a,b) \subset \mathbb{R} : a,b \in \mathbb{Q} \text{ and } a < b\}. \end{align*} We order $P$ by reverse inclusion: \begin{align*} p \leq q \quad \Longleftrightarrow \quad p \subset q. \end{align*} Thus a condition gives information by narrowing the possible real number, so smaller intervals are stronger conditions. The ccc verification is simple because $P$ is countable. The endpoint map sends each interval $(a,b) \in P$ to the ordered pair $(a,b) \in \mathbb{Q}^2$, and $\mathbb{Q}^2$ is countable. Hence $P$ is countable. Every antichain in a countable poset is a subset of a [countable set](/page/Countable%20Set), so every antichain in $P$ is countable. Therefore $P$ satisfies the countable chain condition. [/guided] [/step] [step:Build dense sets for avoidance and shrinking] For each $i \in I$, define \begin{align*} D_i := \{p \in P : \overline{p} \cap F_i = \varnothing\}, \end{align*} where $\overline{p}$ denotes the closure of $p$ in $\mathbb{R}$. We claim that $D_i$ is [dense](/page/Dense%20Subset) in the poset $P$. Let $q = (a,b) \in P$. Since $F_i$ has empty interior, the open interval $q$ is not contained in $F_i$. Hence $q \setminus F_i$ is a non-empty open subset of $\mathbb{R}$, because $F_i$ is closed. Choose $x \in q \setminus F_i$. Since $q \setminus F_i$ is open, there exists $\varepsilon > 0$ such that \begin{align*} (x-\varepsilon,x+\varepsilon) \subset q \setminus F_i. \end{align*} Choose rational numbers $c,d \in \mathbb{Q}$ such that \begin{align*} x-\varepsilon < c < x < d < x+\varepsilon. \end{align*} Then $p := (c,d)$ belongs to $P$, satisfies $p \subset q$, and has \begin{align*} \overline{p} = [c,d] \subset (x-\varepsilon,x+\varepsilon) \subset \mathbb{R} \setminus F_i. \end{align*} Thus $p \leq q$ and $p \in D_i$, proving that $D_i$ is [dense](/page/Dense%20Subset) in $P$. Let $\mathbb{N}:=\{1,2,3,\dots\}$ denote the set of positive integers. For each $m \in \mathbb{N}$, define \begin{align*} E_m := \{p = (a,b) \in P : b-a < 1/m\}. \end{align*} Each $E_m$ is dense in $P$: if $q=(a,b)\in P$, choose rational numbers $c,d$ with $a<c<d<b$ and $d-c<1/m$; then $(c,d)\le q$ and $(c,d)\in E_m$. [guided] The forcing conditions are intervals, so the two things we need from a generic filter are exactly the two things encoded by the sets $D_i$ and $E_m$. First fix $i \in I$. The set \begin{align*} D_i := \{p \in P : \overline{p} \cap F_i = \varnothing\} \end{align*} asks the filter to contain at least one interval whose closed rational hull misses $F_i$. To prove density, start with an arbitrary condition $q=(a,b)\in P$. Since $F_i$ has empty interior, the open interval $q$ cannot be contained in $F_i$. Therefore there is a point $x\in q\setminus F_i$. Because $F_i$ is closed, its complement $\mathbb{R}\setminus F_i$ is open; because $q$ is open, the intersection $q\setminus F_i$ is open. Hence there exists $\varepsilon>0$ with \begin{align*} (x-\varepsilon,x+\varepsilon)\subset q\setminus F_i. \end{align*} The rational endpoints are needed because $P$ was chosen to be countable. Using density of $\mathbb{Q}$ in $\mathbb{R}$, choose $c,d\in\mathbb{Q}$ such that \begin{align*} x-\varepsilon<c<x<d<x+\varepsilon. \end{align*} Then $p=(c,d)$ is a condition in $P$, and $p\le q$ because $p\subset q$. Moreover \begin{align*} \overline{p}=[c,d]\subset (x-\varepsilon,x+\varepsilon)\subset \mathbb{R}\setminus F_i, \end{align*} so $\overline{p}\cap F_i=\varnothing$. Thus $p\in D_i$, and $D_i$ is dense. Now fix $m\in\mathbb{N}$ and define \begin{align*} E_m:=\{p=(a,b)\in P:b-a<1/m\}. \end{align*} This dense set forces the filter to contain intervals of diameter below $1/m$. Given any $q=(a,b)\in P$, density of $\mathbb{Q}$ lets us choose rational $c,d$ with $a<c<d<b$ and $d-c<1/m$. Then $(c,d)\le q$ and $(c,d)\in E_m$. Hence each $E_m$ is dense. [/guided] [/step] [step:Apply $MA_\kappa$ to obtain a filter meeting all dense requirements] Let \begin{align*} \mathcal{D} := \{D_i : i \in I\} \cup \{E_m : m \in \mathbb{N}\}. \end{align*} Since $|I|\leq \kappa$ and $\kappa$ is infinite, we have $|\mathcal{D}|\leq \kappa$. The poset $P$ is ccc, and every member of $\mathcal{D}$ is dense in $P$. By $MA_\kappa$, there exists a filter $G \subset P$ such that \begin{align*} G \cap D_i \neq \varnothing \quad \text{for every } i \in I \end{align*} and \begin{align*} G \cap E_m \neq \varnothing \quad \text{for every } m \in \mathbb{N}. \end{align*} Here filter means that $G$ is upward closed with respect to $\leq$ and directed: for every $p,q\in G$, there exists $r\in G$ such that $r\leq p$ and $r\leq q$. [guided] We now package all requirements into one family of dense sets. Define \begin{align*} \mathcal{D} := \{D_i : i \in I\} \cup \{E_m : m \in \mathbb{N}\}. \end{align*} The family has size at most $\kappa$: the first part has cardinality at most $|I| \leq \kappa$, the second part is countable, and $\kappa$ is infinite. The hypothesis $MA_\kappa$ applies to every poset satisfying the countable chain condition together with at most $\kappa$ many [dense subsets](/page/Dense%20Subset). We have already verified that $P$ is ccc, and the previous step proved that each $D_i$ and each $E_m$ is dense in $P$. Hence $MA_\kappa$ gives a filter $G \subset P$ meeting every dense set in $\mathcal{D}$. Concretely, \begin{align*} G \cap D_i \neq \varnothing \quad \text{for every } i \in I \end{align*} and \begin{align*} G \cap E_m \neq \varnothing \quad \text{for every } m \in \mathbb{N}. \end{align*} We will use both parts: meeting $D_i$ gives avoidance of $F_i$, and meeting all $E_m$ gives arbitrarily short intervals. The directedness clause in the definition of filter will be used to make these interval requirements compatible. [/guided] [/step] [step:Extract a unique real from the directed family of shrinking intervals] For each $p=(a,b)\in G$, let $\overline{p}=[a,b]$ be its closure in $\mathbb{R}$. We prove that \begin{align*} \bigcap_{p\in G}\overline{p} \end{align*} contains exactly one point. First, the family $(\overline{p})_{p\in G}$ has the finite intersection property. Indeed, if $p_1,\dots,p_n\in G$, directedness of $G$ gives, by induction on $n$, an element $r\in G$ such that $r\leq p_j$ for every $j\in\{1,\dots,n\}$. Since the order is reverse inclusion, $r\subset p_j$ for every $j$, hence \begin{align*} \overline{r}\subset \bigcap_{j=1}^{n}\overline{p_j}. \end{align*} The interval $r$ is non-empty, so the finite intersection is non-empty. Since $G\cap E_1\neq\varnothing$, the filter $G$ is non-empty; choose $p_0=(a_0,b_0)\in G$. Every $p\in G$ has a common strengthening $r\in G$ with $r\leq p$ and $r\leq p_0$, so $\overline{r}\subset \overline{p}\cap\overline{p_0}$. Thus the closed subsets $\overline{p}\cap\overline{p_0}$ of the compact interval $\overline{p_0}$ have the finite intersection property. By the [Heine-Borel Theorem](/theorems/5), closed bounded intervals in $\mathbb{R}$ are compact. By the finite-intersection characterization of compactness, their total intersection is non-empty. Therefore there exists \begin{align*} x \in \bigcap_{p\in G}\overline{p}. \end{align*} To prove uniqueness, let $x,y\in\bigcap_{p\in G}\overline{p}$. Fix $m\in\mathbb{N}$. Since $G\cap E_m\neq\varnothing$, choose $p_m=(a_m,b_m)\in G\cap E_m$. Then $x,y\in\overline{p_m}=[a_m,b_m]$, so \begin{align*} |x-y|\leq b_m-a_m<1/m. \end{align*} Since this holds for every $m\in\mathbb{N}$, we get $|x-y|=0$, hence $x=y$. [guided] The filter $G$ is not a sequence and need not be linearly ordered, so we cannot simply say that the intervals are nested. The substitute is directedness. For each $p=(a,b)\in G$, let $\overline{p}=[a,b]$ be its closure in $\mathbb{R}$. First we prove the finite intersection property. Given finitely many conditions $p_1,\dots,p_n\in G$, directedness gives a common strengthening $r\in G$ with $r\leq p_j$ for every $j\in\{1,\dots,n\}$. Because the order is reverse inclusion, $r\subset p_j$ for each $j$, and therefore \begin{align*} \overline{r}\subset \bigcap_{j=1}^{n}\overline{p_j}. \end{align*} Since $r$ is a non-empty interval, this finite intersection is non-empty. To get an actual point in the total intersection, first note that $G$ is non-empty because $G\cap E_1\neq\varnothing$. Fix one condition $p_0=(a_0,b_0)\in G$. For every $p\in G$, directedness gives $r\in G$ with $r\leq p$ and $r\leq p_0$, hence $\overline{r}\subset \overline{p}\cap\overline{p_0}$. Thus the closed subsets $\overline{p}\cap\overline{p_0}$ of $\overline{p_0}$ have the finite intersection property. By the [Heine-Borel Theorem](/theorems/5), the closed bounded interval $\overline{p_0}$ is compact. By the finite-intersection characterization of compactness, the total intersection is non-empty, so there exists \begin{align*} x \in \bigcap_{p\in G}\overline{p}. \end{align*} It remains to show uniqueness. Let $x,y\in\bigcap_{p\in G}\overline{p}$. Fix $m\in\mathbb{N}$. Since $G$ meets $E_m$, choose $p_m=(a_m,b_m)\in G\cap E_m$. Both $x$ and $y$ lie in $\overline{p_m}=[a_m,b_m]$, so their distance is bounded by the length of that interval: \begin{align*} |x-y|\leq b_m-a_m<1/m. \end{align*} This holds for every positive integer $m$. Hence $|x-y|=0$, and therefore $x=y$. [/guided] [/step] [step:Use the avoidance requirements to keep the limit point outside every nowhere dense set] Let $x$ denote the unique point of \begin{align*} \bigcap_{p\in G}\overline{p}. \end{align*} Fix $i\in I$. Since $G\cap D_i\neq\varnothing$, choose $p_i\in G\cap D_i$. By definition of $D_i$, \begin{align*} \overline{p_i}\cap F_i=\varnothing. \end{align*} Since $x\in\overline{p_i}$, it follows that $x\notin F_i$. Because $i\in I$ was arbitrary, \begin{align*} x\notin \bigcup_{i\in I}F_i. \end{align*} As $A_i\subset F_i$ for every $i\in I$, we also have \begin{align*} x\notin \bigcup_{i\in I}A_i. \end{align*} Therefore $\mathbb{R}$ is not the union of the family $(A_i)_{i\in I}$, completing the proof. [guided] Let $x$ be the unique point in \begin{align*} \bigcap_{p\in G}\overline{p}. \end{align*} We now use the avoidance dense sets. Fix $i\in I$. Since $G$ meets $D_i$, choose $p_i\in G\cap D_i$. By the definition of $D_i$, the closed interval attached to $p_i$ misses $F_i$: \begin{align*} \overline{p_i}\cap F_i=\varnothing. \end{align*} But $x$ lies in $\overline{p}$ for every $p\in G$, and in particular $x\in\overline{p_i}$. Therefore $x\notin F_i$. Because $i\in I$ was arbitrary, $x$ avoids every $F_i$: \begin{align*} x\notin \bigcup_{i\in I}F_i. \end{align*} Finally, each original nowhere dense set satisfies $A_i\subset F_i$, so the same point also avoids every $A_i$: \begin{align*} x\notin \bigcup_{i\in I}A_i. \end{align*} Thus the family $(A_i)_{i\in I}$ cannot cover $\mathbb{R}$. [/guided] [/step]