# Prove any subnet of a convergent net is convergent

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

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 $$x$$.