Problem 12178
(American Mathematical Monthly, Vol.127, April 2020) Proposed by S. Portnoy (USA).
Given any functionf : R → R, show that there is a real number x and a sequence x1, x2, . . . of distinct real numbers such thatxn→ x and f (xn) → f (x) as n → ∞.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let {qi}i∈N+ be an enumeration of Q. For any i, n ∈ N+, let Ji,nbe the subset of N+such that j ∈ Ji,n if and only if
qi−1
n, qi+ 1 n
×
qj− 1
n, qj+1 n
∩ {(x, f (x) : x ∈ R} 6= ∅
i. e. the intersection of the open square centered at (qi, qj) and of side 2/n, and the graph of f is non-empty.
Since Q is dense in R, it follows that each set Ji,nis non-empty. For any i, n ∈ N+and any j ∈ Ji,n
let xi,j,n∈ R such that
(xi,j,n, f (xi,j,n)) ∈
qi−1
n, qi+1 n
×
qj−1
n, qj+ 1 n
,
and let X be the set of all those points xi,j,n. It is evident that X is infinite-countable.
Let x ∈ R \ X (which is non-empty since R is uncountable). Then, by the density of Q, there are two sequences (qin)n and (qjn)n in Q such that
(x, f (x)) ∈
qin−1
n, qin+ 1 n
×
qjn−1
n, qjn+ 1 n
.
Therefore jn∈ Jin,n and, by the triangle inequality,
(x, f (x)) ∈
xin,jn,n− 2
n, xin,jn,n+ 1 n
×
f (xin,jn,n) −2
n, f (xin,jn,n) + 1 n
.
Hence xin,jn,n → x and f (xin,jn,n) → f (x) as n → ∞. Since x 6∈ X, there is a subsequence of
(xin,jn,n)n of distinct real numbers.
Remark. Basically we used the fact that the metric space (R2, d) where d is the Euclidean distance is separable (Q2 is countable and dense in R2) and therefore the metric subspace (G, d) is separable too where G := {(x, f (x)) : x ∈ R}.
