Suppose we are asked to prove that the quotient space $\mathbb R/\mathbb Z$ of $\mathbb R$ equipped with the quotient topology is compact. Has this question provided enough information for us to answer it? Do we not need to know the topology given to $\mathbb R$? I ask this because we can attach a wide variety of topologies to $\mathbb R$.
Does this question provide enough information?
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general-topology
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3You may assume that $\Bbb R$ has the usual (Euclidean) topology unless some other topology is specified. – 2012-05-05
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0Yes, it is. If $\Bbb R$ were given the discrete topology, for instance, the quotient would be discrete and infinite and therefore not compact. – 2012-05-05
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0@BrianM.Scott: Thank you. – 2012-05-05
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1@Brian: Perhaps you should write your comment as an answer? – 2012-05-06
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1@mixedmath: Done. (I too like to get questions off the Unanswered list.) – 2012-05-06
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0I'd like a little clarification, as there are two possible meanings of $\mathbb R / \mathbb Z$. One is that this term means the quotient group of the additive group $\mathbb R$ by the additive subgroup $\mathbb Z$; this group is often called the circle group, and is isomorphic to the multiplicative group of complex numbers of modules 1, and is compact. The other is that the symbol $X/A$ means $X$ with $A$ shrunk to a point; in this case the result looks non compact! – 2012-05-13
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You may generally assume that any $\Bbb R^n$ has the usual (Euclidean) topology unless some other topology is explicitly specified or made very clear by the context.