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Can anybody tell me an example of a connected topological space which every convergent sequence in this space is constant (after a finite number of terms)?

thnks!

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A space with only one point has that property. –  MJD Apr 2 at 7:48
    
@MJD Which is arguably an uninteresting example. –  k.stm Apr 2 at 7:49
    
That is why I did not post it as an answer. –  MJD Apr 2 at 7:54

2 Answers 2

up vote 8 down vote accepted

Any uncountable space $X$ with the cocountable topology should do.

Obviously, such a space cannot be written as a disjoint union of two proper closed (i.e. countable) subsets.

On the other hand, if $(x_n)_{n ∈ ℕ}$ is a converging sequence in $X$ with limit $x ∈ X$, then it has to eventually reach the open neighbourhood $U = X\setminus\{x_n;\; n ∈ ℕ, x_n ≠ x\}$ of $x$ in $X$, so from then on, it has to be constantly $x$ (as the only $x_n$ allowed in $U$ are those who satisfy $x_n = x$).

So any convergent sequence is eventually constant there.

This space is not Hausdorff!

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Let $\mathcal{O}_{\text{e}}$ denote the usual Euclidean topology on $\mathbb{R}$, and consider the family $$\mathcal{B} := \{ U \setminus A : U \in \mathcal{O}_{\text{e}} , A \subseteq \mathbb{R}\text{ is countable} \}.$$ Then $\mathcal{B}$ is a base for a topology on $\mathbb{R}$ which is finer than the usual topology. (In the sequel, unless otherwise indicated, topological terminology (e.g., open sets, closed sets) refer to this new topology.)

  • Connected. Suppose that $U,V$ are disjoint nonempty open subsets which cover $\mathbb{R}$. Note that we may write $$U = \bigcup_{i \in I} ( U_i \setminus A_i ); \quad V = \bigcup_{i \in I} ( V_i \setminus B_i )$$ where each $U_i, V_j$ is open in the Euclidean topology, and each $A_i, B_j$ is countable. It follows that $U^\prime = \bigcup_{i \in I} U_i$ and $V^\prime = \bigcup_{j \in J} V_j$ are nonempty open subsets of $\mathbb{R}$ in the Euclidean topology, and $U^\prime \cup V^\prime = \mathbb{R}$. As $\mathbb{R}$ is connected in the Euclidean topology, then $U^\prime \cap V^\prime \neq \varnothing$. Pick $i \in I$, $j \in J$ such that $U_i \cap V_j \neq \varnothing$. Since $U$ and $V$ are disjoint, it follows for each $x \in U_i \cap V_j$ that either $x \in A_i$ or $x \in B_j$. But then one of $A_i$ or $B_j$ is uncountable, which is a contradiction!

  • No nontrivial convergent sequences. Suppose that $\langle x_n \rangle_{n \in \mathbb{N}}$ is a one-to-one sequence in $\mathbb{R}$. Note that $\mathbb{R} \setminus \{ x_n : n \in \mathbb{R} \}$ is an open set which contains no memeber of the sequence, and so the sequence cannot converge to a point in this set. For $m \in \mathbb{N}$, note that $\mathbb{R} \setminus \{ x_n : n > m \}$ is an open neighbourhood of $x_m$ containing no $x_n$ for $n > m$. It follows that the sequence cannot converge to $x_m$.

  • Hausdorff. Since the new topology is finer than the Euclidean topology, this space is clearly Hausdorff.

  • Not separable. It follows that all countable subsets of $\mathbb{R}$ are closed, and hence no countable subset can be dense.

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