# Continuity definitions are equivalent

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## Statement

The definitions of continuity for a function $f:(X,d)\rightarrow(Y,d')$ from one metric space to another is the same as $f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$ being continuous (where the topologies are those induced by the metric are the same, that is

1. $\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))$
2. $\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}$

## Proof

### $\implies$

Suppose $f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$ is continuous.

Let $V\in\mathcal{K}$ - that is $V$ is open within $Y$

Let $x\in f^{-1}(V)$ be given.

Then because $V$ is open, $\exists\epsilon>0$ such that $B_\epsilon(f(x))\subset V$ (note that $f(x)\in V$ by definition of where we choose x from).

But by continuity of $f$ we know that $\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V$

Thus $B_\delta(x)\subset f^{-1}(V)$ (as for all $a$ in the ball, the thing $f$ maps it to is in the ball of radius $\epsilon$ about $f(x)$).

Since $x$ was arbitrary we have $\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)$, thus $f^{-1}(V)$ is open.

### $\impliedby$

Choose any $x\in X$

Let $\epsilon>0$ be given.

As $B_\epsilon(f(x))$ is an open set, the hypothesis implies that $f^{-1}(B_\epsilon(f(x)))$ is open in $X$

Since $x\in f^{-1}(B_\epsilon(f(x)))$ and $f^{-1}(B_\epsilon(f(x)))$ is open, it is a neighborhood to all of its points, that means

$\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))$

Note: we have now shown that $\forall\epsilon>0\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))$

Using the implies and subset relation we see $a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))$

Or just $a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))$

Thus it is continuous at $x$, since $x$ was arbitrary, it is continuous.