I have the following topology homework problem:
Prove that if $Y$ is a subspace of $X$, and $Z$ is a subspace of $Y$, then $Z$ is a subspace of $X$.
My first thought was that clearly, since $Z \subset X$, $Z$ is a subspace with the usual subspace topology. But then there is nothing to show and $Y$ is not used.
Then I thought that maybe you need to show the usual subspace topology on $Z$ from $X$ is the same as the topology on $Z$ from $Y$, given that $Y$ is a subspace of $Z$. With that in mind:
The open sets of $Z$ are of the form $Z \cap O_i$, where $O_i$ is open in $Y$. The open sets of $Y$ are of the form $Y \cap U_i$, where $U_i$ is open in $X$. Thus, the open sets in $Z$ are of the form $Z \cap Y \cap U_i$, $U_i$ open in $X$. Clearly this is not equivalent to the usual subspace topology of $\{Z \cap U_i: U_i$ open in $X$}, unless $Y$ is an open set in $X$, which is not given. Is there something I'm missing?