Is there a non-topos for which the objects having power objects form a topos? Given any category $C$ with finite limits, let $D$ be the full subcategory of $C$ consisting of those objects $X$ for which a power object $P(X)$ exists.
Then, one might ask whether the following properties hold:

*

*$1 \in D$, i.e. $C$ has a subobject classifier.

*If $X,Y \in D$, then also $X \times Y \in D$.

*If $Y \to X$ is a monomorphism and $X \in D$, then also $Y \in D$.

*If $X \in D$, then also $P(X) \in D$.

The above four properties imply that $D$ is a topos. Could this happen without $C$ being itself a topos (so $D \ne C$)?
To show that an object $X$ has a power object, it might help to realize $X$ as a subobject of an object that is known to have a power object. Unfortunately, this is hard to do (for example, we cannot take the "singleton" map $X \to P(X)$ since we don't even know that $P(X)$ exists in the first place).

Let $\kappa$ be any inaccessible cardinal and $\mathsf{Set}_{\leq \kappa}$ the full subcategory of $\mathsf{Set}$ of sets with cardinality less than or equal to $\kappa$, or equivalently of cardinality strictly less than $\kappa^+$. Then $\mathsf{Set}_{\leq \kappa}$ is closed under subobjects, and assuming the axiom of choice is also closed under binary products. It contains the subobject classifier for $\mathbf{Set}$, which is 2. For any object $X$, the power object $\mathcal{P}(X)$ exists precisely when $|2^X| \leq \kappa$, which is exactly when $|X| < \kappa$. However, when $|X| < \kappa$ we in fact have $|2^X| < \kappa$ and so $\mathcal{P}(\mathcal{P}(X))$ also exists.
For example, you can take $\kappa := \omega$. Then $\mathsf{Set}_{\leq \omega}$ is the category of countable sets, and the full subcategory of $X$ where $\mathcal{P}(X)$ exists is the category of finite sets, which is a topos.
