by $ x = (x_1, …, x_n) in mathbb {R} ^ n $, leave $ Q_x = (x_1, x_1 + 1) times … times (x_n, x_n + 1) $ – the open cube you have $ x $ in its "lower left" corner.

Leave $ K subset mathbb {R} ^ n $. We can choose $ v in mathbb {R} ^ n $ so that

For any $ m in mathbb {Z} ^ n $, either $ (K + v) cap Q_m ne varnothing $or $ (K + v) cap overline {Q_m} = varnothing $?

This means that I want to change $ K $ so there is no cube to cross $ K $ only at the limit (of the cube). **If necessary, assume that $ K $ It is compact.** I am pretty sure this should be a well known fact, if true. I don't know what the correct tags are for the question, please help me.

Here is an idea that I am struggling to implement. We need to find $ v = (v_1, .., v_n) in (0,1) ^ n $ so yes $ x = (x_1, .., x_n) in K $ and $ x_k + v_k in mathbb {Z} $so there is no open cube that prevents $ K $ and has $ x $ at its limit so let's consider points of $ K $ that are in the limits of such cubes. We will differentiate ourselves by the size of the face.

For example, for $ n = 2 $ to consider $ F_ {1,1} = {x in K, Q_x cap K = varnothing } $ – the dots in the lower left corner of a square that avoids, $ F_ {1, -1} = {x = (x, y) in K, Q _ {(x, y-1)} cap K = varnothing } $ – the dots in the upper left corner of a square that avoids, $ F -1.1 and $ F -1, -1 $ similarly defined. It is easy to see that these are not dense anywhere.

Also, define $ F_ {1,0} $ be the set of all $ x in mathbb {R} $ such that there is $ y, z in mathbb {R} $ such that $ (x, y) in K $, $ z <y <z + 1 $ and $ Q _ {(x, z)} cap K = varnothing $. This means that $ (x, y) in K $ it is inside the left side of a square that it avoids. $ F -1.0, $ F_ {0,1} $ and $ F_ {0, -1} $ they are defined in a similar way. **The missing step:** shows that the sets $ F_ {1,0} $, $ F -1.0, $ F_ {0,1} $ and $ F_ {0, -1} $ they are small (specifically – lean).

Now the quotient maps $ mathbb {R} a S ^ 1 $ and $ mathbb {R} ^ 2 to (S ^ 1) ^ 2 $ map meager sets to meager sets and (a meager subset of $ S ^ 1 $) times $ S ^ 1 $ is scarce in $ (S ^ 1) ^ 2 $. So, there is $ v = (u, w) in (0,1) ^ 2 $ such that $ x in F _ { pm1,0} Rightarrow x + u not in mathbb {Z} $, $ y in F_ {0, pm 1} Rightarrow y + w not in mathbb {Z} $ and $ (x, y) in F _ { pm1, pm1} Rightarrow (x + u, y + w) not in mathbb {Z} ^ 2 $.

The same idea "works" in the overall dimension module of the missing step (which is probably even more complicated).