Let (x,d) be metric space and A proper subset of X .Define the closure of a set A .consider the usual metric space (Rn,d) .let A = {(x1,x2,.......xn): xi element of Q}
There is three possible definitions of a closure of a set "A" (which are equivalent for a metric space) :
For a space "\\mathbb{R}^n" with a usual metric "d" and a set "A" defined as "x\\in A \\leftrightarrow (x=(x_1,...,x_n) \\wedge \\forall i\\: x_i \\in \\mathbb{Q})" the closure of "A" is the entire space "\\mathbb{R}^n". We can prove it, using the second definition. Let "x\\in\\mathbb{R}^n" with "(x_1,..., x_n)" its components. By density of "\\mathbb{Q}" in "\\mathbb{R}", there are sequences "(x_i^k)_{k\\in \\mathbb{N}}" with "x^k_i \\in \\mathbb{Q}" and "x_i^k\\overset{k\\to+\\infty}{\\to} x_i" for all "1\\leq i\\leq n". The sequence in "A" defined as "((x_1^k,...,x_n^k))_{k\\in\\mathbb{N}}" converges to "x=(x_1,...,x_n)", as all coordinates of "x^k" converge to a respective coordinate of "x". Therefore, "x\\in \\bar A" and as this point is arbitrary, we have "\\bar A=\\mathbb{R}^n".
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