# If $x_n \to a$ and $x'_n \to a$, then $\{x_1, x'_1, x_2, x'_2, …\} \to a$

I know that if all subsequences of $\{x_1, x'_1, x_2, x'_2, ...\}$ converge to $a$, then $\{x_1, x'_1, x_2, x'_2, ...\}$ converges to $a$, but I only know two subsequences of $\{x_1, x'_1, x_2, x'_2, ...\}$ that converge to $a$, namely $x_n$ and $x'_n$. How do I show that there don't exist any other subsequences that converge to something other than $a$?

• for $x_n$ there exists some $n_1$, for $x'_n$, there exists some $n_2$, then for $\{x_1,x'_1,\ldots\}$ take $n_0>2n_1+2n_2$ – Luis Felipe May 10 '15 at 17:25

Let $y=(x_1,x_1',x_2,x_2',...)$. For $n$ odd, we have $y_n = x_{n +1\over 2}$, for $n$ even, we have $y_n = x_{n \over 2}'$.

Let $\epsilon>0$ then there are $N,N'$ such that if $k \ge N$ we have $|a-x_k| < \epsilon$ and if $k \ge N'$ we have $|a-x_k'| < \epsilon$.

Let $M = 2\max(N,N')$ and $n \ge M$. Then if $n$ is odd, we have ${n +1\over 2} \ge N$ and so $|y_n-a| < \epsilon$. Similarly, if $n$ is even, we have ${n \over 2}\ge N'$ and so $|y_n-a| < \epsilon$.

Hence $y_n \to a$.

• Sorry for late +1 but yes, this is exactly what I would have done. It's almost a non-question because "but both get as close as one requires to $a$" – Alec Teal May 10 '15 at 20:12
• @copper.hat I am still having trouble understanding this proof. I don't understand why we need to let $M = 2\max(N, N')$. Say for some fixed $\epsilon$ that $N = 5$ and $N' = 10$. Then if we choose $M = \max(N, N')$, then we choose $N'$ and so since $N' > N$, both sequences converge past when $N'$. So why do we need the extra factor of $2$? – mr eyeglasses Jul 25 '15 at 8:01
• @morphic: The $N,N'$ come from considerations on $x_n,x_n'$, to 'convert' the $N,N'$ for use with $y_n$ you need to multiply by 2. – copper.hat Jul 26 '15 at 6:31
• @copper.hat Does multiplying by 3 also work? – mr eyeglasses Jul 26 '15 at 19:02
• Multiplying the indices by any number $\ge 2$ will work. Remember the sequence is $(y_1,y_2,y_3,y_4,...) = (x_1,x_1', x_2,x_2',...)$. The corresponding indices are $(1,2,3,4,...)$ and $(1,1,2,2,...). – copper.hat Jul 26 '15 at 21:50 Hint: Any convergent sequence is Cauchy. Then notice that every subsequence converging to$b\neq a$implies that the sequence itself converges to$b$, which generates an absurd. Fix$\epsilon>0$and take$N\in\mathbb N$such that for every$n\ge N$,$|x_n-a|<\epsilon$and$|x'_n-a|<\epsilon$. Now observe that for this epsilon the natural number$2N$works. HINT: Use the following lemma: A sequence$\{a_n\}_n$converges to$a$iff every subsequence of$\{a_n\}_n$admits a sub-subsequence that converges to$a$. You can try to prove this lemma by your own, it's not difficult (look at the negation of the definition of convergence). Please notice that the lemma you're trying to use is that a sequence converges to something iff every subsequence of it converges to the same something. The lemma I posted says something different. Here I show you some details: We have to prove that any subsequence of$\{x_1, x'_1, x_2, x'_2, ...\}$contains a sub-subsequence (i.e., a subsequence of the subsequence) that converges to$a$. Take an arbitrary subsequence of$\{x_1, x'_1, x_2, x'_2, ...\}$, two cases: it contains infinitely many elements from the sequence$\{x_n\}_n$or it contains infinitely many elements from the sequence$\{x'_n\}_n$, does this give you an idea about how to get a sub-subsequence that converges to$a\$?