# Notes - Metric Spaces MT23, Limits and continuity

### Flashcards

What does it mean for $x _ n \to x$ in a metric space $X$?

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

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 $a \in X$.

$f$ 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

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

whenever $ \vert \vert x \vert \vert < \delta$. Since $f(0) = 0$, $ \vert \vert f(x) \vert \vert \le 1$ for these $x$. Then if $ \vert \vert v \vert \vert = 1$, $ \vert \vert \delta v / 2 \vert \vert = \delta / 2 < \delta$, so $ \vert \vert f(\delta v / 2) \vert \vert \le 1$.

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

**Bounded implies continuous**: Suppose that $ \vert \vert f(v) \vert \vert < M$ for all $v$ with $ \vert \vert v \vert \vert \le 1$. Fix $\varepsilon > 0$, and let $\delta := \varepsilon / M$. Then if $ \vert \vert v - w \vert \vert < \delta$,

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.