Question about neighboorhoods

analysis

Let $\xi$ be a point in $\mathbb{R}^n$ and $X$ a neighboorhood of $\xi$ and $1 \leq j \leq n$. Define $$I_j(\xi) := \{t \in \mathbb{R} | (\xi_1, \xi_2, ..., \xi_{j - 1}, t, \xi_{j + 1}, ..., \xi_n) \in X\}.$$ Then my book says that it is clear that $I_j(\xi)$ is a neighboorhood of $\xi_j$ because there exists a $\delta > 0$ such that $B(\xi ; \delta) \subset X$. Can anyone prove this?

My attempt: there exists a $\delta > 0$ such that for all $a \in B(\xi, \delta)$ $a \in X$, so $\sqrt{(a_1 - \xi_1)^2 + ... + (a_j - \xi_j)^2 + ... + (a_n - \xi_n)^2} < \delta $ implies that $a \in X$. But I don't get how to link that to $\xi_j$ being in the neighboorhood of $I_j(\xi)$.

Clearly, because $X$ is an open set, there exists some $\delta$ such that $B(\xi, \delta)\subset X$.

Now, take some point $t\in\mathbb R$ such that $|t-\xi_j|<\delta$, and let's define $$\xi_t = (\xi_1,\dots, \xi_{j-1},t,\xi_{j+1}\dots, \xi_n).$$

Then, $$\|\xi - \xi_t\|=\sqrt{(\xi_1-\xi_1)^2 + (\xi_2-\xi_2^2)+\cdots+(\xi_{j-1} - \xi_{j-1})^2 + (\xi_j-t)^2 + (\xi_{j+1}-\xi_{j+1})^2+\cdots + (\xi_n-\xi_n)^2} \\=\sqrt{0+\cdots+0+(\xi_j-t)^2+0+\cdots+0} = |\xi_j-t| < \delta$$

which means that, for every $t$ such that $|t-\xi_j|<\delta$, you have $\|\xi-\xi_t\|<\delta$ which means $\xi_t\in B(\xi, \delta)\subset X$.

Now, this also means that since $\xi_t\in X$, that $t\in I_j(\xi)$. So, you know now that