Let $X$ be a topological space and $Y$ a subset of $X$. Write $i: Y \to X$ for the inclusion map. Choose the correct statement:

  1. If $i$ is continuous, then $Y$ has the subspace topology.

  2. If $Y$ is an open subset of $X$, then $i(U)$ is open in $X$ for all subsets $U \subseteq Y$ that are open in the subspace topology on $Y$.


  1. Since $i$ is continuous means $i^{-1}(V)$ is open in $Y$ for any open $V$ in $X$. So can we say first option is wrong as $Y$ can have any topology which is stronger than the subspace topology.

  2. Since $U$ is open in subspace topology which means there exist an open set $V \in X$ such that $U = V \cap Y$. Then $i(U) = i(V\cap Y) $

    How to think further?

  1. Yes, the first assertion is false, but you should provide an example. For instance, you can take $X=\mathbb R$ with the usual topology and $Y=[-1,1]$ with the discrete topology.
  2. $i(U)=U=V\cap Y$, which is open in $X$, since both $V$ and $Y$ are open in $X$.
  • $\begingroup$ Can you tell me the open set in $Y$ which is not of the form $V \cap Y$ for $V$ open in $X$ for part 1? $\endgroup$ – Mittal G May 1 '18 at 20:10
  • $\begingroup$ @MittalG No, I can't, because I made a mistake. I've edited my answer. I hope that everything is clear now. $\endgroup$ – José Carlos Santos May 1 '18 at 20:20

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