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Let $f:J \rightarrow X$ be a net in $X$, let $f(\alpha) = x_\alpha$. If $K$ is a directed set and $g:K \rightarrow J$ is a function such that $i \leq j \rightarrow g(i) \leq g(j)$ and $g(K)$ is cofinal in $J$, then the composite function $f \circ g: K \rightarrow X$ is called a subnet of $(x_\alpha)$. Prove that if the net $(x_\alpha)$ converges to $x$, so does any subnet.

I'm having a lot of trouble proving this. I know that $f(\alpha)=x_\alpha = x$ means that $\forall U$ neighborhoods of $x, \exists \alpha \in J$ such that $\alpha \leq \beta \rightarrow x_\beta \in U$, i.e. $f(\beta) \in U$ and the desired result is to have this for $f(g(\alpha))$. I'm guessing we now consider an arbitrary sequence in $K$, but I really can't figure it out

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Let $U$ be a neighboorhood of $x$. As $x_\alpha\to x$, there exists an index $\beta\in J$ such that $\alpha\geq\beta \rightarrow x_\alpha\in U$. As the image $g(K)$ is cofinal in $J$, there exists $\theta \in K$ such that $g(\theta)=\beta'\geq \beta$. Then $\zeta\ge\theta\rightarrow g(\zeta)\geq g(\theta)=\beta'\ge\beta$, hence $\zeta \geq\theta \rightarrow f(g(\zeta))=x_{g(\zeta)}\in U$, which means the subnet $f\circ g:K\rightarrow X$ coverges to $x$.

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