Meital Posted April 6, 2005 Posted April 6, 2005 If we have X with order topology, and we have [a,b] subset of X. And if we know that we have U open in X, can we conclude that [a,b] /\ U is open in [a,b]? If so please explain.
Dapthar Posted April 6, 2005 Posted April 6, 2005 If we have X with order topology, and we have [a,b] subset of X. And if we know that we have U open in X, can we conclude that [a,b] /\ U is open in [a,b']? If so please explain.If you're imposing the relative topology (also known as the subspace topology) on [math][a,b][/math], then, by definition, if [math]U[/math] is open in [math]X[/math], then [math]U \cap [a,b][/math] is open in [math][a,b][/math].
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