I am hoping to have my proof reviewed for verification. Thanks.


Let $X$ be a topological space and let $Y$ be a metric space. Let $f_n: X \to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points in $X$ converging to $x \in X$. Show that if the sequence $(f_n)$ converges uniformly to $f$, then $f_n(x_n)$ converges to $f(x)$.


Since each $f_n$ is continuous, $Y$ is a metric space, and $f_n$ converges to $f$ uniformly, then $f$ is continuous.

Let $U$ be an open set containing $f(x)$. Take $V$ to be the open ball of some radius $ε$ centered at $f(x)$, where $V$ is contained within $U$.

Let $W$ be the open ball of radius $\dfracε2$, centered at $f(x)$. Then so far we have: $W \subset V \subset U$.

Then $f^{-1}(W)$ is an open set containing $x$. Then there exists some $n_1$ such that for $n > n_1, x_n \in f^{-1}(W)$. Then $f(x_n) \in W$ for $n > n_1$.

Then choose an $n_2$ such that $d(f(x), f_n(x)) < \dfracε2, \forall x, \forall n > n_2$. Such an $n_2$ exists since $f_n$ converges uniformly to $f$. (Note: $d$ is the distance function on the metric space $Y$).

Take $n_3$ to be maximum of $n_1$ and $n_2$. Then for $n > n_3$, the sequence $f_n(x_n)$ is contained within $V$ and hence, $U$. So the sequence converges to $f(x)$ since $U$ was chosen to be an arbitrary open set containing $f(x)$.


It's a completely valid proof which I would write differently (more formulae, fewer words):

Take $\varepsilon > 0$. As $f$ is continuous at $x$ (for the reasons you stated) we can find an open neighbourhood $U$ of $x$ such that

$$\text{ (1): } \forall p \in U : d(f(p) ,f(x)) < {\varepsilon \over 2}$$

And as $x_n \to x$, we can find $N_1 \in \mathbb{N}$ such that:

$$\text{ (2): } \forall n \ge N_1: x_n \in U$$

Combining (2) with (1) we thus have:

$$\text{ (2a): } \forall n \ge N_1: d(f(x_n), f(x)) < {\varepsilon \over 2}$$

And by uniform convergence of $f_n$ to $f$, we can find $N_2 \in \mathbb{N}$ such that:

$$\text{ (3): } \forall n \ge N_2: \forall x \in X: d(f_n(x), f(x)) < {\varepsilon \over 2}$$

Now taking $n \ge N_3 = \max(N_1, N_2)$, by the triangle inequality we see:

$$d(f_n(x_n), f(x)) \le d(f_n(x_n), f_n(x)) + d(f_n(x), f(x)) < {\varepsilon \over 2} + {\varepsilon \over 2} = \varepsilon$$

because we can apply (2a) and (3) at the end. As $\varepsilon>0$ was arbitrary, we have shown $f_n(x_n) \to f(x)$, as required.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.