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Question:

Let $X$ be a connected space with a topology not necessarily sequential. What is an example where a point in $X$ is not the limit of any not eventually constant sequence?

Motivation. Suppose $X$ is a topological vector space over $\mathbb{R}$ or $\mathbb{C}$. If we define a sequentially separated set $S$ of $X$ as such that, for every $x\in S$, $x$ lies outside the sequential closure of subspace $Y_x:=\text{span}(S\setminus \{x\})$. I'm trying to use the usual Zorn's lemma argument claiming there always exists a maximal such set. But it seems, if $X$ has a not-so-nice topology, there might be points which cannot be approximated by any not eventually constant sequence, and that can derail the reasoning.

I came to think about this issue when trying to understand uncountable Schauder basis. Thanks.

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    $\begingroup$ Do you mean "not the limit of any not eventually constant sequence"? Otherwise, you can start with whatever terms you want, and them become constant. $\endgroup$
    – Xander Henderson ♦
    Commented Feb 15 at 13:23
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    $\begingroup$ In any event, a good approach for problems like this is to consider small examples. I think that the set $\{a,b\}$ with the topology $\{\varnothing, \{a\}, \{a,b\}\}$ does the job. The space is connected and $a$ is not the limit of any sequence which is not eventually constantly $a$. $\endgroup$
    – Xander Henderson ♦
    Commented Feb 15 at 13:24
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    $\begingroup$ Surely, they really mean: "is not a sequential limit point" $\endgroup$
    – FShrike
    Commented Feb 15 at 13:24
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    $\begingroup$ I agree, the Sierpinski space is a very lightweight counterexample. To the author: if you actually only care about the TVS or Banach space situation, you should refocus to this; in general, as written, your question easily has a negative answer $\endgroup$
    – FShrike
    Commented Feb 15 at 13:25
  • $\begingroup$ @XanderHenderson Thanks. I'll updated the question. $\endgroup$
    – user760
    Commented Feb 15 at 13:26

2 Answers 2

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Let $X = \omega_1\times [0, 1) \cup \{\omega_1\}$ where $\omega_1\times [0, 1)$ is given lexicographic order and $\omega_1$ is a point which is greater than all points of $\omega_1\times [0, 1)$.

Give $X$ the order topology.

Then $X$ is connected and $\omega_1$ is not a limit of a not eventually constant sequence.

enter image description here

Imagine $X$ as a long closed interval, modification of the long ray to which we add a top element $\omega_1$, or as the ordinal $\omega_1+1$ to which we fill in the gaps between ordinals using segmemts.

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When looking for pathological examples in topology (or just counterexamples, in general), it is usually best to start with very small, simple cases. Don't make your life too complicated—try to find the smallest space which has the properties you need. In this case, consider a two point set $\{a,b\}$ with the topology $\{\varnothing, \{a\}, \{a,b\}\}$.

  • This space is connected, as there do not exist two nonempty disjoint open sets (let alone two nonempty disjoint open sets whose union is the entire space).

  • Any sequence which has limit $a$ must eventually be constantly $a$.

NB: As FShrike points out in the comments, this space is the Sierpinski space. I've never heard that term before, so I learned something new today. Huzzah.

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    $\begingroup$ Sierpiński space is, up to homeomorphism, the only non-trivial (i.e. not discrete nor indiscrete) topology with two points and it has this very useful property that any continuous function $f:X\to S$ where $S$ is Sierpiński space, corresponds to an open set of $X$. $\endgroup$
    – Jakobian
    Commented Feb 15 at 13:34
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    $\begingroup$ Heh. I learned that term rather recently too. In general I don't like the practice of using proper names as mathematical nomenclature, and this example does not further endear me to that practice. $\endgroup$
    – Lee Mosher
    Commented Feb 15 at 13:38
  • $\begingroup$ Thanks. I'll then have to think about it in the TVS context $\endgroup$
    – user760
    Commented Feb 15 at 13:41
  • $\begingroup$ In a TVS over $\mathbb{R}$ or $\mathbb{C}$, you have a lot more structure. Given $v$ in your TVS, the sequence $\bigl( (1+1/n)v \bigr)_{n}$ converges to $v$ (for example). $\endgroup$
    – Xander Henderson ♦
    Commented Feb 15 at 13:51

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