Suppose A is a subset of a topological space X. How do you show that every subset of A, which is open with respect to the subspace topology on A, is open with respect to the topology on X? The wikipedia page for subspace topology claims this in the 'properties' section.



1 Answer 1


This is not true as written. Suppose that $A\subset X$, where $X$ is a topological space, and $A$ is not open in $X$. But $A$ is open in $A$ when $A$ is equipped with the subspace topology, while it still isn't open in $X$. This gives us a contradiction.

However, the result is true when $A$ is itself open in $X$. This is because if $B\subset A$ is open in $A$, then $B=U\cap A$ where $U$ is open in $X$. But since $U$ and $A$ are both open in $X$ and finite intersections of open sets are open, we find that $B$ is open in $X$ as well.

  • $\begingroup$ Ok that makes sense thanks. Actually the wikipedia page claims that later statement which you show is indeed true. My mistake. $\endgroup$
    – Smithey
    Feb 7, 2015 at 20:04
  • $\begingroup$ Happy to help.. $\endgroup$
    – Hayden
    Feb 7, 2015 at 20:05

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