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Let $X$ be a topological space, and let $\mathcal{A}$ be an open cover for $X$.

To say that $Open(X)$ is coherent with $\mathcal{A}$ means that $$B\in Open(X) \Leftrightarrow B\cap A\in Open(A)\forall A \in \mathcal{A}$$

I am trying to prove that $Open(X)$ is always coherent with $\mathcal{A}$, whenever $\mathcal{A}$ is an open cover of $X$.

(This is from Spanier's Algebraic Topology pp. 5)


However, I get stuck, because as far as I understand it, $Open(A)$ is not necessarily the usual subspace topology from $Open(X)$ (I thought so at first, but then the whole claim seems trivial and nonsensical), but some pre-specified topology, and we merely have to show $Open(X)$ is coherent with all these topologies.


So I start by assuming that $\mathcal{A}\subseteq Open(X)$ and $\bigcup_{A\in\mathcal{A}}A = X$ and, for the $\Longleftarrow$ direction, pick an arbitrary $B\subseteq X$ such that $(B\cap A)\in Open(A) \forall A\in\mathcal{A}$ and try to show that $B\in Open(X)$. But I don't know anything about $Open(A)$, so I don't know how to proceed.

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  • $\begingroup$ This notion of coherence looks very interesting... $\endgroup$ – goblin Sep 9 '18 at 3:15
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You do have to use the subspace topology ; otherwise it might not be true that the topology of $X$ is coherent with $\mathcal{A}$. For example, look at what happens when $\mathcal{A} = \{ X \}$.

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  • $\begingroup$ Then perhaps I misunderstand the construction of the topology coinduced by a collection of maps $g_i:X_i\to X$. I thought it was given by $Open(X) := \{U\subseteq X\,|\,g_i^{-1}(U)\in Open(X_i) \forall i\}$, which, for the case of a collection of subsets of $X$, $\mathcal{A}$, with $g_i$ being the inclusion maps, reduces to $Open(X) = \{U\subseteq X\,|\,U\cap A\in Open(A)\forall A\in\mathcal{A}\}$. But if $Open(A) = \{U\cap A\,|\,U\in Open(X)\}$ is the subspace topology, then we get $Open(X) = \{U\subseteq X\,|\,U\in Open(X)\}$ which means that any collection of subsets of $X$ furnished with $\endgroup$ – PPR May 15 '15 at 13:42
  • $\begingroup$ the subspace topology makes $Open(X)$ coherent with it. So what's the sense of the definition?? $\endgroup$ – PPR May 15 '15 at 13:42
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    $\begingroup$ @PPR No, not any collection of subsets with the subspace topology makes $Open(X)$ coherent. If $U \in Open(X)$, then $U \cap A \in Open(A)$ by definition of the subspace topology. However, the converse is false : if $U$ is a subset of $X$ such that $U \cap A \in Open(A)$, you cannot conclude that $U \in Open(X)$ in general. All you can say is that $U \cap A = V \cap A$ for some $V \in Open(X)$. $\endgroup$ – Math536 May 15 '15 at 20:10
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    $\begingroup$ @PPR For example, take $\mathcal{A}$ to be the set of singletons of $X$. Then $Open(X)$ is coherent with $\mathcal{A}$ if and only if $X$ has the discrete topology. $\endgroup$ – Math536 May 15 '15 at 20:16
  • $\begingroup$ I understand my mistake. Thank you so much. $\endgroup$ – PPR May 15 '15 at 21:52

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