There are closed spaces that are not complete

Question

I am learning functional analysis and over the last week we have been introduced to some topological notions. I have seen that a space is complete when any Cauchy sequence in that space approaches something and that something is inside the space. From here, I could reasonably quickly come up with spaces that are not Cauchy (like (rationals, absolute value) , (irrationals,absolute value) which made me realize that these have in common that the set used is not closed , which lead me to understand that complete spaces are necessarily closed. I didn’t even think about whether the converse statement though since it did not occur natural to me to think about that converse statement. But then I started working on the problem : Is the space $C([0,1])$ with metric induced by the maximum norm a complete space? My intuition was telling me NO since I could think of a sequence lying in the space which seemed to approach a function not in the space which by definition is a function discontinuous somewhere on [0,1]. More specifically my sequence in mind was ${t^n}$. I was initially convinced that I could show that the sequence approached the discontinuous function $F(t) = 0 $if $t$ in $[0,1)$ and $1$ if $t=1$ however when I tried to formalize it I noted that I cannot measure distance from my discontinuous “limit” function to any of the terms of the sequence because the maximum is simply not defined for an increasing function on an open interval. So now this has made me examined my approach which was basically trying to argue that the set is not closed. But it seems like fue to the maximum not being well defined with the type of discontinuous functions that sequences in this set may approach makes me now think that the space is closed! So if it is closed (which I am pretty sure it is) then, the only way for not being complete is that there is some Cauchy sequence which doesn’t even approach anything to begin with. And I think that my given sequence is of this type. Because I cannot think of any continuos function for which the sequence approaches it in the sense of the metric given. I would really like to receive feedback on my thoughts and confirmation that they are on the right track… If they are, that is: I am right thinking that the space is closed but not complete, how can I show that the sequence cannot approach any continuos function and thus not be a complete space formally? Thank you so much in advance!

Answer 1

You have to be a bit careful when you say that a space is closed. Say you live in a normed (or even metric) space $X$. Then $X$ will always be a closed set in it self. However, when speaking about a subset $A \subset X$ you can indeed examine when it is closed.

You were correct to notice that if $A \subset X$ is complete, then it is closed in $X$. This is somewhat trivial: If $(x_n)$ is a sequence in $A$ converging to some point $x \in X$ then $(x_n)$ is Cauchy in $A$ so that it converges to some point $a \in A$. By uniqueness of the limit you have that $x=a \in A$ so that the set $A$ is (sequentially) closed.

The converse is not true in general, unless $X$ is complete. Then you have the equivalence $$A \text{ is closed in $X$} \iff A \text{ is complete }$$ To see why the converse direction is true in this case, consider a closed set $A \subset X$ in a complete space $X$. Then any a Cauchy sequence in $A$ is also Cauchy in $X$ so that it converges to some point $x \in X$ due to completeness of $X$. But $A$ is closed so the limit point $x$ must lie in $A$. Hence any Cauchy sequence in $A$ converges in $A$.

What you are trying to show about $C[0,1]$ is not true since this is indeed a complete metric space. If you were to endow $C[0,1]$ with the $L^2$ norm $\|f\|_2 = \left ( \int_0^1 |f(x)|^2 \,dx \right)^{1/2}$ then $(C[0,1], \|\cdot \|_2)$ is not a complete space. You can show this directly by constructing a $\|\cdot\|_2$-Cauhcy sequence in $C[0,1]$ that does not converge in $C[0,1]$ or indirectly by showing that $C[0,1]$ is a subset of the space of square integrable functions $L^2(0,1)$ - which is complete - and show that $C[0,1]$ is not closed in $L^2(0,1)$ with the $L^2$ norm.