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Consider the function $f: \Bbb R^+ \to \Bbb R$ given by $f(x) = \log(x)$. Then $f$ is continuous. The domain of $f$ is $\Bbb R^+$ and the range of $f$ is $\Bbb R$.

We have the following fact: (proven here)

A function $g : X \to Y$ is continuous if and only if for every closed $W \subseteq Y$, its preimage $g^{-1}(W) \subseteq X$ is closed.

But $f$ is continuous, $\Bbb R$ is closed and its preimage $\Bbb R^+ = (0, +\infty)$ is open. This seems to be a contradiction. I assume I'm missing something obvious...what's going on here?

aras
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    $\mathbb{R}^+$ is closed inside itself- – mathma May 27 '17 at 17:19
  • @mathma Sorry, I don't see why that is? The sequence $a_n = 1/n$ converges to $0$, and $a_n \in \Bbb R^+$ but $0 \not\in \Bbb R^+$. – aras May 27 '17 at 17:23
  • @aras Like you said $0 \not\in \mathbb{R}^{+}$, so $a_n \to 0$ doesn't make sense. If you were asking whether $\mathbb{R}^{+}$ was closed inside $\mathbb{R}$ then your sequence would make sense as $1/n \to 0 \in \mathbb{R}$. – Zain Patel May 27 '17 at 17:27
  • Isn't the set of real numbers both open and closed? Perhaps that would affect the validity of applying the rule. – Paul Aljabar May 27 '17 at 17:31
  • In short, $a_n=\frac1n$ is not a convergent sequence in $\mathbb R^+$. (This shows that $\mathbb R^+$ is not a complete metric space.) –  May 27 '17 at 17:32
  • You need to consider the relative topology when using that theorem. Closed sets in $\mathbb{R}^+$ are of the shape $F\cap\mathbb{R}^+$, where $F$ is closed in $\mathbb{R}$. For instance (0,1] would be closed there but not in $\mathbb{R}$. And in particular $\mathbb{R}^{+}$ is closed. – mathma May 27 '17 at 17:33
  • @mathma That makes sense, thanks! – aras May 27 '17 at 17:34
  • The open ends are not actually achievable by the function as the input values to achieve are not in the domain. The domain of valid arguments is open so the domain of valid outputs is (at least) half open. – CogitoErgoCogitoSum May 27 '17 at 17:48

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The domain $\mathbb{R}^+$ needs to be considered as its own topological space and not as a subset of $\mathbb{R}$.

If you do this, then $\mathbb{R}^+$ is closed because any sequence of points in $\mathbb{R}^+$ that converges, does so to a point of $\mathbb{R}^+$.

Thompson
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