Suppose that $f:\mathbb{Q}\to\mathbb{R}$ is a uniformly continuous function. For all $x\in\mathbb{R}$, $x$ is a cluster point of $\mathbb{Q}$. Show that $\displaystyle\lim_{y\to x}f(y)$ is defined for all $y\in\mathbb{R}$.
I have provided an answer using proof by contradiction. Please correct if my answer does not make sense anywhere. Also, how would I answer using a direct proof? Thanks.
Suppose $f:\mathbb{Q}\to\mathbb{R}$ is a uniformly continuous function on $\mathbb{Q}$, and for any $x\in\mathbb{R}$, $x$ is a cluster point of $\mathbb{Q}$. Now, suppose that $\displaystyle \lim_{y \to x}f(y)$ is not defined for all $y\in \mathbb{R}$. Then, according to the Divergence Criteria, the sequnce $(y_n)$ in $\mathbb{Q}$, with $y_n\neq x$ for all $n\in\mathbb{N}$, is such that $(y_n)$ converges to $x$, and the sequence $(f(y_n))$ does not converge in $\mathbb{R}$. Since the sequence $(y_n)$ in $\mathbb{Q}$, with $y_n\neq x$ for all $n\in\mathbb{N}$, is convergent, then according to the Cauchy Convergence Criterion, $(y_n)$ is a Cauchy sequence. Thus, according to Part (1), $(f(y_n))$ is a Cauchy sequence, so that according to the Cauchy Convergence Criterion, the sequence $(f(y_n))$ converges in $\mathbb{R}$. However, the sequence $(f(y_n))$ does not converge in $\mathbb{R}$. Thus, a contradiction has been reached. Therefore, $\displaystyle \lim_{y \to x}f(y)$ is defined for all $y\in \mathbb{R}$.
Part (1) refers to the proof that if $f:\mathbb{Q}\to\mathbb{R}$ is a uniformly continuous function and $(x_n)$ is a Cauchy sequence in $\mathbb{Q}$, then $(f(x_n))$ is a Cauchy sequence in $\mathbb{R}$.