Hi:(adsbygoogle = window.adsbygoogle || []).push({});

Just curious: a continuous function f:X-->Y ; X,Y topological spaces, can fail

to be an embedding because it is not 1-1, or, if f is 1-1 , f can fail to be an

embedding because, for U open in X f(U) is not open in f(X).

Can anyone think of a "reasonable" example of the last case, by reasonable

I mean no spaces with finitely many points, the inclusion map from a space

into itself with a different topology (i.e.: we have (X,T) and (X,T') , and we

use i:X-->X : i(x)=x ) , or maybe or something one could find in

"Counterexamples" book.

If A is a strict subspace of X , then the subspace topology on A guarantees

that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.?

Thanks.

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# Continuous bijection that is not an embedding

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