In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence?

This is a cross-posted on MSE here.

Let $$(X,d)$$ be a metric space. Say that $$x_n\in X$$ is a P-sequence if $$\lim_{n\rightarrow\infty}d(x_n,y)$$ converges for every $$y\in X.$$ Say that $$(X,d)$$ is P-complete if every P-sequence converges. Problem 1133 of the College Mathematics Journal (proposed by Kirk Madsen, solved by Eugene Herman) asks you to prove that $$\text{compact}\Longrightarrow\text{P-complete}\Longrightarrow\text{complete}$$ and that none of these implications go both ways. The implications follow by showing that $$\text{sequence}\Longleftarrow\text{P-sequence}\Longleftarrow\text{Cauchy sequence},$$ since a P-sequence (and thus a Cauchy sequence) converges iff it has a convergent subsequence. To give counterexamples to the converses, there are several possible directions. My question specifically involves normed vector spaces (although it is overkill for the original problem).

For any $$n\geq 0$$, any norm on $$\mathbb R^n$$ induces a P-complete metric. This distinguishes compactness and P-completeness, since $$\mathbb R^n$$ obviously isn't compact when $$n>0$$. To differentiate P-completeness and completeness, we can note that a Hilbert space is P-complete iff it is finite-dimensional (otherwise, we take a non-repeating sequence of vectors from an orthonormal basis and get a P-sequence that doesn't converge). I wonder if other infinite-dimensional normed spaces (necessarily Banach) might be P-complete. But my knowledge of Banach spaces is very limited, so I don't have much intuition about what examples to try. Also, the property of P-completeness (unlike compactness and completeness) is not closed-hereditary, so we can't just try an something by embedding it in a larger example.

Question: What is an example of an infinite dimensional, P-complete Banach space?

Examples I've tried:

• $$\ell^p$$ spaces for all $$1\leq p< \infty$$. They are not P-complete, since the sequence $$e_n=(0,\dotsc,0,1,0,\dotsc)$$ is a P-sequence but not Cauchy. As was pointed out to me in the comments, $$\ell^\infty$$ is not P-complete, but you need a different sequence as a counterexample.
• $$C(X)$$ for $$X$$ compact Hausdorff, first-countable and infinite. There must be an accumulation point $$p\in X$$. We can take a sequence of bump functions $$f_k$$ converging (pointwise) to the characteristic function $$\chi_p$$. For any $$g\in C(X)$$, we have $$\lim d(g,f_k)=\lVert g-\chi_p\rVert_\infty$$. Thus $$(f_k)$$ is a P-sequence that does not converge (uniformly), because the pointwise limit is discontinuous.
• This is a nice question! I hope you don't mind that I highlighted the question itself, and added the requirement of infinite dimensionality, which it seemed to me from context to be very likely that you meant to impose. (Also, I noticed that you ran into spacing issues with $||g - \chi_p||_\infty$ ||g - \chi_p||_\infty, which you addressed with $||g - \chi_p{||}_\infty$ ||g - \chi_p{||}_\infty. Spacing and semantics can be achieved with $\|g - \chi_p\|_\infty$ \|g - \chi_p\|_\infty or, for maximal explicitness, $\lVert g - \chi_p\rVert_\infty$ \lVert g - \chi_p\rVert_\infty.) – LSpice Jul 23 at 20:47
• Awesome! Thanks so much. – Nikhil Sahoo Jul 23 at 20:48
• $e_n$ is not a $P$ sequence as a sequence from $\ell^{\infty}$ (take $y=(0,-1,0,-1,\ldots)$). – Christian Remling Jul 23 at 22:45
• What seems to work though in this case is something like $x_n=\chi_{[n,\infty)}$ (this is similar in spirit to your second example). – Christian Remling Jul 23 at 23:30

That every Banach space is contained in a $$P$$-complete Banach space follows immediately from the following

Theorem. Let $$X$$ be a Banach space. Then there exists a Banach space $$Y$$ containing $$X$$ in which no separated sequence is a $$P$$-sequence.

Modulo "abstract nonsense", which I will explain later, the theorem follows from the following proposition, which comes from Christian Remling's remark that the unit vector basis $$(e_n)$$ of $$c_0$$ is not a $$P$$-sequence in $$\ell_\infty$$.

Proposition. Suppose that $$(x_n)$$ is a normalized basic sequence in a Banach space $$X$$. Then there is an isometric embedding $$S$$ from $$X$$ into $$X \oplus_\infty \ell_\infty$$ such that no subsequence of $$(Sx_n)$$ is a $$P$$-sequence.

Proof: Since $$(x_n)$$ is normalized and basic and $$\ell_\infty$$ is $$1$$-injective, there is $$\alpha >0$$ and a contraction $$T: X \to \ell_\infty$$ such that for all $$n$$, $$Tx_n = \alpha e_n$$. Define $$S$$ from $$X$$ into $$X \oplus_\infty \ell_\infty$$ by $$Sx := (x,Tx)$$. Since $$T$$ is a contraction, $$S$$ is an isometric embedding. We show that $$(Sx_n)$$ does not contain a $$P$$-convergent subsequence; this is basically Christian's comment. Let $$A$$ be any infinite set of natural numbers and take an infinite subset $$B$$ of $$A$$ so that $$A\setminus B$$ is also infinite. Then the distance from $$Sx_n$$ to $$-1_B$$ is $$1+\alpha$$ if $$n$$ is in $$B$$ and one otherwise, so $$(x_n)_{n\in A}$$ is not a $$P$$-sequence.

Now comes the soft souping up. By iterating the Proposition transfinitely, we get for any Banach space $$X$$ a superspace $$Z$$ such that no normalized basic sequence in $$X$$ is a $$P$$-sequence in $$Z$$. Iterate this $$\omega_1$$ times to get an increasing transfinite sequence $$X_\lambda$$, $$\lambda < \omega_1$$, of Banach spaces with $$X_1 = X$$ so that no normalized basic sequence in $$X_\lambda$$ is a $$P$$-sequence in $$X_{\lambda+1}$$. Let $$Y$$ be the union of $$X_\lambda$$ over $$\lambda < \omega_1$$. Every sequence in $$Y$$ is in some $$X_\lambda$$, hence no normalized basic sequence in $$Y$$ is a $$P$$-sequence. This property carries over to the completion of $$Y$$ by the principle of small perturbations.

Now suppose that $$Y$$ is a Banach space in which no normalized basic sequence is a $$P$$-sequence. We claim that also no separated sequence in $$Y$$ is a $$P$$-sequence. Certainly no non norm null basic sequence in $$Y$$ is a $$P$$-sequence, and $$P$$-sequences are bounded, so it is enough to consider a general separated sequence $$(x_n)$$ that is bounded and bounded away from zero. If the sequence has a basic subsequence, we are done. But it is known (and contained, for example, in the book of Albiac and Kalton), that if such an $$(x_n)$$ has no basic subsequence then it has a subsequence that converges weakly, so without loss of generality we can assume that $$x_n - x$$ converges weakly to zero but is bounded and bounded away from zero. But then $$x_n - x$$ has a basic subsequence, hence $$x_n - x$$ cannot have a $$P$$-subsequence, whence neither can $$x_n$$.

EDIT 7/27/20: The reduction of the problem to the theorem above is a consequence of things proved, but perhaps not always explicitly stated, in any course that contains an introduction to metric spaces:

Theorem. Let $$M$$ be a metric spaces. Then one and only one of the following is true.

A. $$M$$ is totally bounded.

B. $$M$$ contains a separated sequence.

A corollary is that every sequence in a metric space either contains a Cauchy subsequence or a separated subsequence.

• Thanks, this is great! I feel like I somewhat understand most of the argument (sans the more advanced details). Since this is not an area in which I know a lot, I want to check that I actually understand the immediate implication that you mention in the beginning. Does the following sound right? – Nikhil Sahoo Jul 27 at 20:03
• Let $(x_n)$ be a $P$-sequence. To show $P$-completeness, we want $(x_n)$ to have a Cauchy subsequence. Let $d_m=\lim_{n\rightarrow \infty}d(x_m,x_n)$. Since $P$-sequences and separated sequences are preserved under taking subsequences, the assumption is that $(x_n)$ has no separated subsequences and thus $\lim_{m\rightarrow \infty}d_m=0.$ Define $n_k$ as follows. Pick $n_0$ so that $d_{n_0}<1.$ If $n_k$ is such that $d_{n_k}<2^{-k}$, we can pick $n_{k+1}>n_k$ with $d_{n_{k+1}}<2^{-(k+1)}$ and $||x_{n_k}-x_{n_{k+1}}||<2^{1-k}$. Then $(x_{n_k})$ is Cauchy by the triangle inequality? – Nikhil Sahoo Jul 27 at 20:18
• Yes, but I would explain this as a consequence of facts that are standard for a course that covers metric spaces. I'll add to the answer. – Bill Johnson Jul 27 at 22:45
• Ah yes. Thank you, this is very instructive and a much cleaner statement. – Nikhil Sahoo Jul 27 at 23:19

It seems to me that you can show that no infinite-dimensional separable Banach space $$X$$ is P-complete as follows. Pick any bounded separated sequence $$\{x_n\}_{n=1}^\infty$$ in $$X$$ and pick a dense sequence $$\{y_i\}$$ in $$X$$. Pick a subsequence in $$\{x_n\}$$ for which $$\|x_n-y_1\|$$ converges. Then from this subsequence pick further subsequence for which $$\|x_n-y_2\|$$ converges. So on. After doing this for all $$i$$, pick a diagonal subsequence $$\{x_{n(k)}\}_{k=1}^\infty$$ and show that it satisfies the desired conditions.

• Thanks! For the past two days, I've been grasping at ideas regarding subsequences of bounded sequences, so it feels affirming to know that I was on the right track. Even if there may still be some non-separable example out there, I'm quite happy with this answer to the negative in the separable case, since sequential conditions on topological spaces tend not to work as nicely without some sort of countability condition. – Nikhil Sahoo Jul 24 at 6:34
• This is of course a very good answer -- although it does not answer the question as stated. Perhaps it should not have been accepted so quickly. – Jochen Wengenroth Jul 24 at 10:16
• @NikhilSahoo You are welcome. One can extend this argument to some classes of nonseparable spaces. It can happen that the general nonseparable case is difficult. – Mikhail Ostrovskii Jul 24 at 22:55