Why we need complete space for the interchange of limits to be valid?

Question

I am learning Real analysis using the book Analysis 2 by Terence Tao. I'm confused with the condition on which we can interchange the limit.

Concerning the below proposition, my question is that: Why we need the space Y to be complete ? What will happen if the space Y is not complete ?

Thank you for your help!

Answer 1

Consider the following counterexample where $Y$ is not complete.

Let $X = [0,1]$ and $Y = (0,2)$. Let $E = (0,1)$. Consider the sequence $f^{(n)}:E\to Y$ with $f^{(n)}(x) := x+1/n$ and $f(x) := x$. Let $x_0 = 0$.

Then

• $\lim_{x\to 0}f^{(n)}(x)=\frac1n$ for each $n$;
• $f^{(n)}$ converges to $f$ uniformly on $E$.

But the limit of the sequence $(\frac1n)_{n=1}^\infty$ does not exist in $Y$.

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In the conclusion of the proposition, it is said that the limit of the sequence $$ (\lim_{x\to x_0;x\in E}f^{(n)}(x))_{n=1}^\infty $$ exists, which is not an assumption.

One way to prove it is to show that it is a Cauchy sequence using the following estimate (for $x\in E$ near $x_0$ and large $n,m$): $$ d_Y(g(x_0;n),g(x_0;m)) =d_Y(g(x_0;n),f^{(n)}(x))+d_Y(f^{(n)}(x),f^{(m)}(x))+d_Y(f^{(m)}(x),g(x_0;m)) $$ where $ g(x_0;n):=\lim_{x\to x_0;x\in E}f^{(n)}(x) $.

This is where you need completeness of $Y$. If $Y$ is not complete, then the limit cannot be guaranteed to exist.