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I'd like to show whether the intersection of two ideals is again an ideal or not. For this, consider two ideals $\mathfrak{h_1},\,\mathfrak{h_2}$ of the Lie algebra $\mathfrak{g}$. In order to answer this question I simply want to check whether the definition of ideals i.e. $[\mathfrak{h_1}\cap\mathfrak{h_2},\mathfrak{g}]\subseteq\mathfrak{h_1}\cap\mathfrak{h_2}$ is fullfilled or not. But I don't know how to express an arbitrary element of $\mathfrak{h_1}\cap\mathfrak{h_2}$.

I guess that the bracket is a subset of the intersection $[\mathfrak{h_1},\mathfrak{h_2}]\subseteq\mathfrak{h_1}\cap\mathfrak{h_2}$. For this bracket, the condition $[[\mathfrak{h_1},\mathfrak{h_2}],\mathfrak{g}]\subseteq [\mathfrak{h_1},\mathfrak{h_2}]$ is easy to show (using the Jacobi identity), but this is lacking the required generality.

How can I write down an arbitrary element of $\mathfrak{h_1}\cap\mathfrak{h_2}$? Additionally the same issue arises when I want to check whether the union $\mathfrak{h_1}\cup\mathfrak{h_2}$ is an ideal or not.

Juri V
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    There's no need to "write down" an arbitrary element of $\mathfrak{h_1}\cap\mathfrak{h_2}$. When you need to say something about such an element you simply say "suppose $x\in \mathfrak{h_1}\cap\mathfrak{h_2}$". (That means that $x\in \mathfrak{h_1}$ and $x\in\mathfrak{h_2}$.) – David C. Ullrich Jun 14 '21 at 17:15
  • @David Ah well, of course! Doing it this way makes the argument very simple, I guess I thought "too simple" in the beginning and then wanted to do it a unnecessarily complicated way. Thanks! – Juri V Jun 14 '21 at 17:27
  • The proof goes the same way as for rings. Just write it down with elements. – Dietrich Burde Jun 14 '21 at 19:18

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