Problem: Let $(X,d_{1})$ and $(Y,d_{2})$ be metric spaces. Suppose that $A$ is a dense subset of X, $X\xrightarrow{f} Y$ is continuous, and $f$ is uniformly continuous when restricted to $A$. Prove that $f$ is uniformly continuous.

My attempt: Suppose all of the above assumptions. Let $\varepsilon>0.$ Fix $\delta>0$ so that, for all $a,b\in A, $ $d_2(fa,fb)<\varepsilon/3$ when $ d_{1}(a,b)<\delta$. Now let $x,y\in X$ satisfy $d_{1}(x,y)<\delta$. We’ll show that $d_{2}(fx,fy)<\varepsilon.$ Because A is a dense subset of metric space $X$, there must exist $(x_n)$,$(y_n)\in$ $A^{\mathbb{Z^{+}}}$ which converge to $x$ and $y$, respectively. By assumption of continuity, $(fx_n)$ and $(fy_n)$ converge to $fx$ and $fy$ respectively. So there must exist $m_1$ and $m_2$ such that, for all $n_i>m_i,$ $d_2(fx_{n_1},fx),d_2(fy,fy_{n_2})<\varepsilon/3$. Choose $n>maxm_i$ so that $d_1(x_n,y_n)<\delta$. Then by construction of $\delta$, $d_2(fx,fy)<d_2(fx,fx_n)+d_2(fx_n,fy_n)+d_2(fy_n,fy)<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon.$ Therefore $f$ is uniformly continuous. $\blacksquare$


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