Relative interior of Cartesian product

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I need to manipulate a relative interior in the context of a convex optimisation problem I am dealing with. However, it is a notion I am not familiar with and I would like to check that my reasoning is correct. I want to prove $-$ disclaimer: I am unsure whether this is true $-$ the following property:

$$ \forall \, i \in [1,k], \, n_i \in \mathbb{N} \, : \, \text{relint} \left(\prod_{i=1}^k \mathbb{R}^{n_i}\right) = \prod_{i=1}^k \mathbb{R}^{n_i} $$

By commutativity of the Cartesian product and the relative interior (Relative interior commutes with cartesian product), it suffices to prove for $n \in \mathbb{N}$:

$$\text{relint}(\mathbb{R}^n) = \mathbb{R}^n$$

Now, letting $\text{aff}$ be the affine hull of a set $S$, to me it is clear that:

$$ \text{aff}(\mathbb{R}^n) = \mathbb{R}^n $$

Indeed, from the definition of an affine hull:

$$ \text{aff}(\mathbb{R}^n) = \left \{ \sum_{i=1}^m \lambda_i x_i : m>0, x_i \in \mathbb{R}^n, \lambda_i \in \mathbb{R}, \sum_{i=1}^m \lambda_i = 1 \right \} $$

It suffices to take $m=1$ and $\lambda \equiv \lambda_1 = 1$, then every element $x$ of $\mathbb{R}^n$ is equal to $\lambda x$.

Subsequently, from the definition of the relative interior $-$ where $N_{\epsilon}(x)$ is a ball of centre $x$ and radius $\epsilon$:

$$\text{relint}(\mathbb{R}^n) = \left \{ x : x \in \mathbb{R}^n, \ \exists \, \epsilon \in \mathbb{R}_+^*, \ N_{\epsilon}(x) \cap \text{aff}(\mathbb{R}^n) \subseteq \mathbb{R}^n \right \} $$

Given that $\text{aff}(\mathbb{R}^n) = \mathbb{R}^n$ and for any $x \in \mathbb{R}^n$, any ball $N_{\epsilon}(x)$ is in $\mathbb{R}^n$, we get:

$$\ N_{\epsilon}(x) \cap \text{aff}(\mathbb{R}^n) \subseteq \mathbb{R}^n \Leftrightarrow \ N_{\epsilon}(x) \subseteq \mathbb{R}^n $$

which is true for all $x \in \mathbb{R}^n$, and we can conclude that:

$$ \text{relint}(\mathbb{R}^n) = \mathbb{R}^n $$

Is my reasoning correct?

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Your proof is correct, but in general if the interior of a set is nonempty then interior and relative interior coincide i.e.,

$$ \text{int} (A) = \text{relint} (A) $$

Now your claim trivially follows .