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I am trying to solve the following exercise:

Let $\Lambda$ be a directed set, and for each $\alpha \in \Lambda$ let $\Gamma_\alpha$ be a directed set. Suppose that for each $\alpha \in \Lambda$ there is a net $({x_k}^{\alpha})_{k \in \Gamma_\alpha}$ that converges to $x^{\alpha} \in X$, and such that $(x^{\alpha})_{\alpha \in \Lambda}$ converges to $x \in X$. Consider $\Phi=\Lambda \times \prod_{\alpha \in \Lambda} \Gamma_{\alpha}$ ordered by the product, which means

$$ (\alpha, (k_\beta)_{\beta \in \Lambda}) \geq (\alpha', (k_\beta')_{\beta \in \Lambda}) \space \text{if and only if} \space \alpha \geq \alpha' \space \text{and} \space k_{\beta}\geq k_{\beta}' \space \forall \beta \in \Lambda$$

Then the net $(\alpha, (k_\beta)_{\beta \in \Lambda}) \rightarrow {x_{k_{\alpha}}}^{\alpha}$ converges to $x$. $$$$$$$$ I am trying to understand the net they ask me to prove it converges, $\alpha$ determines that $(\alpha, (k_\beta)_{\beta \in \Lambda})$ will be send to an element of the net $\Gamma_{\alpha}$. Now, I am not so sure what the subscript $k_{\alpha}$ determines. I don't completely get how the map $(\alpha, (k_\beta)_{\beta \in \Lambda}) \rightarrow {x_{k_{\alpha}}}^{\alpha}$ works, I would appreciate if someone could explain me that and a hint to show the statement is true. Thanks in advance

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Assuming you already understand that the set $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$ with the given ordering is directed, then the proposed net (in the space $X$) works like this. To be a net in $X$, it has to associate, to each element of the index set, some point in $X$. So look at an element of the index set $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$; it's a pair consisting of an element $\alpha\in\Lambda$ and an element of $\prod_{\alpha\in\Lambda}\Gamma_\alpha$. The latter is, by definition of products, a system $(k_\beta)_{\beta\in\Lambda}$ indexed by $\Lambda$, such that, for each $\beta\in\Lambda$, the element $k_\beta$ with index $\beta$ is in $\Gamma_\beta$. Given such a pair $(\alpha,(k_\beta)_{\beta\in\Lambda})$, the net ignores most of the system $(k_\beta)_{\beta\in\Lambda}$, and pays attention only to the one element in this system with index $\alpha$, namely $k_\alpha$. In other words, it uses the first component $\alpha$ of the pair $(\alpha,(k_\beta)_{\beta\in\Lambda})$ to select one element $k_\alpha$ from the second component. Then it goes back to the originally given system of nets $(x_k^\alpha)_{k\in\Gamma_\alpha}$, looks at the $\alpha$-th of these nets, and picks out the $k_\alpha$-term from that net.

To show that this complicated-looking net converges to $x$, I'd consider an arbitrary open neighborhood $U$ of $x$ and see what I can prove to be in $U$. Since the net $(x^\alpha)_{\alpha\in\Lambda}$ converges to $x$, there must be some $\alpha'\in\Lambda$ such that $x^\alpha\in U$ for all $\alpha\geq\alpha'$. Then for each such $\alpha$, the fact that $(x^\alpha_k)_{k\in\Gamma_\alpha}$ converges to a point $x^\alpha$ in the open set $U$ means that there must be some $k'_\alpha$ such that $x^\alpha_k\in U$ for all $k\geq k'_\alpha$ in $\Gamma_\alpha$. I've almost defined an element $(\alpha',(k'_\alpha)_{\alpha\in\Lambda})$ of $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$, for which I hope that all larger elements are mapped into $U$ by the complicated-looking net. There are two catches: (1) I've defined $k'_\alpha$ only for $\alpha\geq\alpha'$; what about other $\alpha$'s in $\Lambda$? (2) I've only hoped, not proved, that all larger elements of $\Lambda\times\prod_{\alpha\in\Lambda}\Gamma_\alpha$ get mapped into $U$. My answer to (1) is extremely cheap: choose the remaining $k'_\alpha$'s, for $\alpha\not\geq\alpha'$, arbitrarily in the corresponding $\Gamma_\alpha$'s. My answer to (2) is even cheaper: I'll leave that to you because I've already done too much work.

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