The subsequential limits of a sequence {$p_n$} in a metric space $X$ form a closed subset of $X$ (Rudin)

Question

Here is the above theorem's proof:

The question:

Why isn't $d(q,p_{n_i})$ strictly less than $2^{1-i}\delta$, since the triangular inequality says:

$$d(q,p_{n_i}) \le d(x,q) + d(x,p_{n_i})<2(2^{-i}\delta)=2^{1-i}\delta \ ?$$

Am I wrong? Or am I unusually picky?

Answer 1

You are correct, since the earlier inequalities are strict, despite the triangle inequality not being strict.

However, note that the statement, as made, is correct(although weaker), and leads to the same conclusion. It does not matter : if $x < \epsilon$ for every $\epsilon > 0$, or if $x \leq \epsilon$ for every $\epsilon > 0$, it still holds that $x=0$.

You should not blame yourself for being picky, though!