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So I am beginning Munkres' textbook on topology.

The topological definition of continuity reads:

$f:X\rightarrow Y$ is continuous if for each open subset $V\subset Y$, $f^{-1}(V)$ is an open subset of $X$.

Of course, it does fit the epsilon-delta definition of continuity since in


$|x-x_0|<\delta$ and $|f(x)-f(x_0)|<\epsilon$ are both open.

Also, since openess of a set means closeness of that set's complement, the following epsilon-delta definition (which is valid) also agree with the topological definition:


Yet my question is, how about the following epsilon-delta definition?


1.) Is it a valid definition of continuity of $\mathbb{R}^n$ regardless of the topological definition?

2.) If yes, does it agree with the topological definition of continuity?

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Note that if $\forall\epsilon\exists\delta,|x-x_0|\le\delta\Rightarrow|f(x)-f(x_0)|<\epsilon$ then $\forall\epsilon\exists\delta,|x-x_0|\le\delta\Rightarrow|f(x)-f(x_0)|\le \epsilon $ – mfl May 27 '14 at 21:39
@ManuelFdzLpz $|x−x_0|≤δ$ is not open, but $|f(x)−f(x_0)|<\epsilon$ is. What about that? – Sanath K. Devalapurkar May 27 '14 at 21:39
Consider $\mathbb{R}$ with the topology where $(-\infty,a)$ is open for any $a.$ Then $id:\mathbb{R}\rightarrow\mathbb{R}$ is continuous. If we define continuity by the condition that the image of "small" closed neighbourhoods lies in an open nhood, then $id$ is no longer continuous. If we consider $x=0$ then $[0,\infty)$ is a closed nhood of $0$ but the only open set that contains $id( [0,\infty))$ is $\mathbb{R}.$ "We would expect" it must be contained in the open nhood $(-\infty,\epsilon>0)$. In other words, the definitions wouldn't be equivalent. – mfl May 27 '14 at 22:42

1 Answer 1

You can consider 4 variations of the $\epsilon$-$\delta$ definition for continuity at a point $x_0$ in $\mathbb{R}^n$.

  1. $\forall \epsilon>0 : \exists \delta > 0: \forall x: (| x - x_0 | < \delta \Rightarrow |f(x) - f(x_0)| < \epsilon)$.
  2. $\forall \epsilon>0 : \exists \delta > 0: \forall x: (| x - x_0 | \le \delta \Rightarrow |f(x) - f(x_0)| < \epsilon)$.
  3. $\forall \epsilon>0 : \exists \delta > 0: \forall x: (| x - x_0 | < \delta \Rightarrow |f(x) - f(x_0)| \le \epsilon)$.
  4. $\forall \epsilon>0 : \exists \delta > 0: \forall x: (| x - x_0 | \le \delta \Rightarrow |f(x) - f(x_0)| \le \epsilon)$.

One can easily check that all of these are mutually equivalent, and we can even do this in any pair of metric spaces instead of $\mathbb{R}^n$. But this is a simple consequence of properties of the order in $\mathbb{R}$ and properties of $\mathbb{R}$ itself.

E.g. to see that 1 implies 2: assume 1 holds, and pick $\epsilon>0$. Then 1 guarantees us a $\delta>0$ with the properties of 1. and then for 2. we just pick $\delta' = \frac{\delta}{2} > 0$. Then if $|x - x_0| \le \delta' = \frac{\delta}{2} < \delta$, so 1 guarantees that indeed $|f(x) - f(x_0)| < \epsilon$.

Other implications go similarly.

And all of these are equivalent to the topological definition. The proof of which uses the first, the most standard, variant as this corresponds best to interior points / open balls.

(Proof in one direction: suppose $O$ is open in the image, and let $x_0$ be in $f^{-1}[O]$. Then $f(x_0) \in O$, so there is some open ball with radius $\epsilon>0$ such that $B(f(x_0), \epsilon) \subset O$. Apply definition 1 to this $\epsilon$ to get $\delta>0$, and then one checks that $B(x_0, \delta) \subset f^{-1}[O]$. The $<$ signs give us nice open balls, so the proof is "smooth")

But the $\le$ instead of $<$ does not have anything to do with inverse images of closed sets being closed etc. The 4 variants are mutually equivalent, and the first one is the more natural to prove the equivalence to the topological definition.

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how do we know that $f^{-1}[O]$ is open? from the beginning of your proof we just know that $x_{0} \in f^{-1}[O]$ ? – arcolombo Oct 1 '14 at 16:27
@AnthonyColombo Because the $\delta$ shows that $x_0$, which is an arbitrary point of $f^{-1}[O]$, is an interior point of $f^{-1}[O]$. So all points of $f^{-1}[O]$ are interior points. – Henno Brandsma Oct 1 '14 at 16:35

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