Is $\mathbb R$ endowed with countable complement topology path connected ? I only know that it is connected . Please help . Thanks in advance
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No. Note that non-singleton, countable sets in the co-countable topology are disconnected. Indeed, let $E$ be at most countable and non-singleton (and non-empty), and let $x \in E$. Then both $\{x\}$ and $E \setminus \{x\}$ are open, disconnecting $E$.
Now, let $a,b \in \mathbb R$, $a \neq b$. Suppose there exists $f : [0,1] \to \mathbb R$ which is continuous from the Euclidean topology to the co-countable topology and has $f(0)=a$ and $f(1)=b$. Let $D \subset [0,1]$ be a countable, dense subset. Since $f([0,1])$ is connected, and is non-singleton since it contains both $a$ and $b$, it is uncountable. Therefore $f([0,1]) \setminus f(D)$ is non-empty, and since $f(D)$ is at most countable, it is open. Thus $f^{-1}(f([0,1]) \setminus f(D)) \subset [0,1] \setminus D$ is non-empty and open. But this contradicts the density of $D$.
Justthisguy
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The last part of the proof can be also said like this. We have $f([0, 1]) = f(\overline{D}) ⊆ \overline{f(D)} = f(D)$. So the image of $f$ is countable, and hence discrete. Since $[0, 1]$ is connected, $f([0, 1])$ is a singleton. – Adam Bartoš Aug 09 '16 at 14:38
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It is still true for cofinite topology or not – Gob Aug 09 '16 at 15:46
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@Gob: It is not. Every injective map from $[0, 1]$ to $ℝ$ with cofinite topology is continuous. So the space is path connected, but it is not arcwise connected since it does not contain a homeomorphic copy of $[0, 1]$. – Adam Bartoš Aug 09 '16 at 17:09
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An easy teenage version approach:
As @Justthisguy has explained, $f([0,1])$ is uncountable. Since $[0,1]$ is compact in Euclidian topology, $f([0,1])$ is compact in co-countable topology. But compact sets in co-countable topology are finite, contradiction.
moqui
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