Two questions: Construction of sequences and Subsequence I have two questions regarding sequences: 1. I was asked to construct a sequence such that for any given $n \in \mathbf{N}$, the sequence has a sub-sequence converging to $n$. 2. Let $x_n$ be a sequence such that the set $x_n$ is finite. Show that there is a sub-sequence $x_{n_i}$ such that $\forall i, x_{n_i}=x_{n_1}$. For the first question I know that the sequence is not convergent, but it is however bounded. I was not able to come up with a proper construction.

The example given in the comments by J. W. Tanner - 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,... etc would definitely work since it has a sub sequence - 1, 2, 3, 4, .... n, n, n, n, ... which converges to n.

You said that the sequence is bounded. But since n can be **any** natural number, you would need to "approach" it i.e. "come near" n if it is converging to n. Hence it can't be bounded.

For the second question, try proving by contradiction. We know that the set ${x_n}$ is finite. This means that the sequence can only take finitely many values. Now, suppose there is no sub-sequence $x_{n_i}$ such that $ ∀i,x_{n_i}=x_{n_1}$. This means there is no sub-sequence of ${x_n}$ which is constant. So no term is repeated infinitely many times ( to form a sub-sequence). But since there are finite options for each term, being repeated finitely many times each would give a finite no. of terms. This contradicts the sequence having infinite terms. Hence there has to be a constant sub-sequence.