Exercise 3.3.1 (Uniform limits preserve continuity I)

Question

Prove Proposition 3.3.1.\par 

	extbf{Proposition 3.3.1} Suppose $\left(f^{(n)}ight)_{n=1}^{\infty}$ is a sequence of functions from one metric space $\left(X, d_{X}ight)$ to another $\left(Y, d_{Y}ight)$, and suppose that this sequence converges uniformly to another function $f: X ightarrow Y$. Let $x_{0}$ be a point in $X$. If the functions $f^{(n)}$ are continuous at $x_{0}$ for each $n$, then the limiting function $f$ is also continuous at $x_{0}$.

Prakash Anand · Accepted Answer

\begin{proof} Since $f^{(n)}$ converges uniformly to $f$, we know that given $\epsilon^{\prime}>0$ there exists $N>0$ such that $d_{Y}\left(f^{(n)}(x), f(x) ight)<\epsilon$ for all $n>N$ and $x \in X$. Since $f^{(n)}$ is continuous at $x_{0} \in X$ we know that given $\epsilon^{\prime \prime}>0$ there exists $\delta^{\prime}>0$ such that if $d_{X}\left(x, x_{0} ight)<\delta^{\prime}$ then $d_{Y}\left(f^{(n)}(x), f^{(n)}\left(x_{0} ight) ight)<\epsilon^{\prime \prime}$ So given $\epsilon>0$ take $\epsilon^{\prime}=\epsilon / 3, \epsilon^{\prime \prime}=\epsilon / 3$ and $\delta=\delta^{\prime}$. If $d_{X}\left(x, x_{0} ight)<\delta=\delta^{\prime}$, then $$ \begin{aligned} d_{Y}\left(f(x), f\left(x_{0} ight) ight) & \leq d_{Y}\left(f(x), f^{(n)}(x) ight)+d_{Y}\left(f^{(n)}(x), f^{(n)}\left(x_{0} ight) ight)+d_{Y}\left(f^{(n)}\left(x_{0} ight), f\left(x_{0} ight) ight) \ &<\epsilon / 3+\epsilon / 3+\epsilon / 3 ext { (by uniform convergence and continuity) } \ &=\epsilon \end{aligned} $$ Thus $f$ is continuous at $x_{0}$. \end{proof}

Exercise 3.3.1 (Uniform limits preserve continuity I)

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Exercise 3.3.1 (Uniform limits preserve continuity I)

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