# Question About Proving $\alpha(\cdot)$ is Continuous in the Proof of Proposition 3.C.1 from MWG

See Proposition 3.C.1 from MWG

Continue from this post, Microeconomic Theory by Mas-Colell et al. book said the following:

What remains is to establish that all convergent subsequences of $$\{\alpha(x^n)\}_{n=1}^{\infty}$$ converge to $$\alpha(x)$$.

The book then proceed to prove this and conclude:

Since all convergent subsequences of $$\{\alpha(x^n)\}_{n=1}^{\infty}$$ must converge to $$\alpha(x)$$, we have $$\lim_{n\to\infty}\alpha(x^n)=\alpha(x)$$, and we are done.

I seriously think this is wrong. Consider the sequence $$0,1,0,2,0,3,0,4,0,5,\dots$$. Every convergent subsequence of this sequence converges to 0, but this sequence diverges.

We need to prove every subsequence (not just the convergent ones) of $$\alpha(x^n)$$ converges to $$\alpha(x)$$ (although this idea is not helpful because the original sequence is a subsequence of itself).

What am I missing here? How should we fix this? Thank you very much!

Here is the full proof of the final step of Proposition 3.C.1 from MWG:

• Please keep including explicit references (as edited by me in your last two posts and now this one) in subsequent posts instead of referring to "the book". Thank you! Commented May 13 at 19:12
• @RichardHardy Actually, the "MWG" stands for the book Microeconomic Theory. Commented May 13 at 19:49
• Acronyms are helpful when they are introduced properly. That ensures the reader is on the same page as the writer and hardly costs anything to include. Failing to introduce an acronym may often be a sign of sloppiness, though some may perceive it as disrespect for the reader, too. Commented May 14 at 5:37

The sequence you proposed $$0,1,0,2,0,3,\ldots$$ i.e. $$x_n=\begin{cases}0 &\text{if n is odd} \\ \frac{n}{2} & \text{if n is even} \end{cases}$$ is not a bounded sequence and is therefore, not from a compact set. The following result holds for sequences from compact sets:
In a compact metric space $$(X,d)$$, if every convergent subsequence of a sequence converges to the same limit, say $$l$$, then the original sequence also converges to $$l$$. [ Ref: See this result- https://math.stackexchange.com/q/461610/378131 ]