# Notes - Metric Spaces MT23, Limits and continuity

> Source: https://ollybritton.com/notes/uni/part-a/mt23/metric-spaces/notes/limits-and-continuity/ · Updated: 2023-07-20 · Tags: uni, notes

- [Course - Metric Spaces MT23](https://ollybritton.com/notes/uni/part-a/mt23/metric-spaces/)

### Flashcards
What does it mean for $x_n \to x$ in a metric space $X$?::

$$
\forall \varepsilon > 0 \text{ } \exists N \in \mathbb N \text{ s.t. } \forall n > N \text{ } d(x_n, x) < \varepsilon
$$

Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces. What does it mean for $f : X \to Y$ to be continuous?::

$$
\forall \varepsilon > 0 \text{ } \exists \delta > 0 \text{ s.t. } \forall x \in X \text{ } d_X(x, a) < \delta \implies d_Y(f(x), f(a)) < \varepsilon
$$

Can you give the characterisation of the continuity at a single point of $f : X \to Y$ about $a \in X$ in terms of neighbourhoods?::

$f$ is continuous at $a \in X$ if and only if for every neighbourhood $N \subseteq Y$ of $f(a)$, the preimage $f^{-1}(N)$ is a neighbourhood of $a \in X$.

Quickly prove the equivalence between the $\varepsilon$-$\delta$ characterisation of continuity at a point $a \in X$ and the neighbourhood characterisation, i.e.

> $f$ is continuous at $a \in X$ if and only if for every neighbourhood $N \subseteq Y$ of $f(a)$, the preimage $f^{-1}(N)$ is a neighbourhood of $a \in X$.

::

**Continuous implies neighbourhood property**: Fix some neighbourhood $N$ of $f(a)$. Then $\exists \varepsilon > 0$ such that $B(f(a), \varepsilon) \subseteq N$. By the $\varepsilon$-$\delta$ characterisation of continuity, there exists $\delta$ such that if $x \in B(a, \delta)$, then $f(x) \in B(f(a), \varepsilon)$. Then
$$
f^{-1}(N) \supseteq f^{-1}(B(f(a), \varepsilon)) \supseteq B(a, \delta)
$$
Since $B(a, \delta)$ is a neighbourhood of $a$, we are done.

**Neighbourhood property implies continuous**: Fix $\varepsilon > 0$. Then $B(f(a), \varepsilon)$ is a neighbourhood of $f(a)$. By assumption, $f^{-1}(B(f(a), \varepsilon))$ is a neighbourhood of $a$, so contains a ball $B(a, \delta)$. In other words, $x \in B(a, \delta) \implies f(x) \in B(f(a), \varepsilon)$. We are done.

Can you give the characterisation of continuity of $f : X \to Y$ in terms of open sets?::

$f$ is continuous on all of $X$ if and only if for each open subset $U$ of $Y$, its preimage $f^{-1}(U)$ is open in $X$.

Can you give an example of a linear map which is not continuous in a metric space?::

Consider $X = \\{f : [-1, 1] \to \mathbb R : f \text{ differentiable}\\}$ with the $\sup$ norm. Then take $F : X \to \mathbb R$ given by $F(f) = f'(0)$. Then consider $f_n(x) = \frac{1}{n} \sin(nx)$.

Suppose:

- $V, W$ are normed vector spaces
- $f : V \to W$ is linear

Quickly show that

$$
f \text{ continuous} \iff \\{||f(x)|| : ||x|| < 1\\}
$$
is bounded.::

**Continuous implies bounded**: $f$ is continuous at $0 \in V$, so take $\varepsilon = 1$ in the definition of continuity. Then $\exists \delta > 0$ such that
$$
d(f(x), f(0)) < 1
$$
whenever $||x|| < \delta$. Since $f(0) = 0$, $||f(x)|| \le 1$ for these $x$. Then if $||v|| = 1$, $||\delta v / 2|| = \delta / 2 < \delta$, so $||f(\delta v / 2)|| \le 1$.

Then $||f(\delta v / 2)|| = \delta ||f(v)|| / 2$. Then $||f(v)|| \le 2/\delta$. So $\\{||f(x)|| \mid ||x|| \le 1\\}$ is bounded.

**Bounded implies continuous**: Suppose that $||f(v)|| < M$ for all $v$ with $||v|| \le 1$. Fix $\varepsilon > 0$, and let $\delta := \varepsilon / M$. Then if $||v - w|| < \delta$,
$$
\begin{aligned}
||f(v) - f(w)|| &= ||f(v-w)|| \\\\
&= \delta ||f(\delta^{-1}(v -w))|| \\\\
&< \delta M \\\\
&= \varepsilon
\end{aligned}
$$
as required.

### Proofs
Let $f : X \to Y$ be a function. Prove that
$$
f \text{ continuous at } a \iff \forall (x_n)^\infty_{n=1} \text{ s.t. } \lim_{n \to \infty} f(x_n) = f(a)
$$
::

Todo.

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