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Find an example of a topological space with a topology different from the indiscrete topology in which every subset is connected.

I can only think of a space with one single point, but then every topology will be the indiscrete topology.

  • For a general characterization of these spaces, see https://math.stackexchange.com/questions/1890361/a-topological-space-such-that-every-subspace-from-it-is-connected – Eric Wofsey Apr 18 '17 at 20:19

2 Answers2

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If $A = \{ a,b \}$ with topology $\tau = \{ \varnothing, \{a\}, A \}$, then it is not the indiscrete topology.
Isn't this one connected?

Edit:
This is the Sierpiński space.
And it is indeed connected.

amrsa
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I think $X = \mathbb{N}$ with the topology $\{\emptyset ,\mathbb{N}\} \cup \{L_n: n \in \mathbb{N}\}$ where $L_n =\{1,\ldots,n\}$, the so-called lower topology.

The same set with the upper topology $\{\emptyset, \mathbb{N}\} \cup \{U_n: n \in \mathbb{N}\}$, where $U_n = \{k \in \mathbb{N}: k \ge n\}$ also has this property.

Henno Brandsma
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  • In a way, the Sierpinski space is a particular case of your first example: if we think of $\mathbb{N}$ in your example as an ordinal, and in the Sierpinski space it's the finite ordinal $2$. I suspect this can even be generalized for other ordinals. Certainly for finite ones, maybe for limit ordinals, at least. – amrsa Apr 19 '17 at 07:52
  • I think Eric Wofsey's answer to the question in the link he gives in a comment to the question answers my suspicion in the positive. – amrsa Apr 19 '17 at 07:58
  • @amrsa yes, the open sets being linearly ordered is crucial for these examples. – Henno Brandsma Apr 19 '17 at 15:29