3

We proved in a lecture that the metric continuity at a point is equivalent to the topological continuity at that point:

A function $f\colon (X_1, d_1)\to (X_2, d_2)$ is continuous at $x\in X_1$ iff for every open ball $B$ centered around $f(x)$, there exists an open ball around $x$ that is mapped inside $B$ via $f$.

A function $f\colon (X_1, \tau_1)\to (X_2, \tau_2)$ is continuous at $x\in X_1$ iff for every open neighborhood $V$ of $f(x)$, we have that $f^{-1}(V)$ is open.

But then, consider $f\colon \mathbb R\to \mathbb R$ (both under the standard metric topology, so that the above result applies) given by $$ f(x) := \begin{cases} x, & x\ne -1\\ 1, & x = -1 \end{cases}. $$ Then $f$ is clearly metric continuous at $1$. But it's not topological continuous at $1$ since $(0, 2)$ is an open neighborhood of $f(1)$ and $f^{-1}((0, 2)) = (0, 2)\cup\{-1\}$ which is not open.

What is wrong with this "counterexample"?

It turns out that my definition for topological continuity is not quite "correct". The correct one is the one that Zerox proposed in the comments: $f$ is continuous at $x$ iff for every (open) neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood (not necessarily open) of $x$. This implies that for every open neighborhood $V$ of $f(x)$, there exists an open neighborhood $U$ of $x$ such that $f(U)\subseteq V$.

The confusion crept in since I though that the above implies my definition for topological continuity, which is not the case. My "proof" went like this:

Let $V$ be an open neighborhood of $f(x)$. To show that $f^{-1}(V)$ is open, let $y\in f^{-1}(V)$, i.e., $f(y)\in V$. Since $V$ is an open neighborhood for $f(y)$ too, we can choose an open neighborhood $U$ of $y$ such that $f(U)\subseteq V$. Then $U\subseteq f^{-1}(f(U))\subseteq f^{-1}(V)$.

Then Zerox pointed out the flaw in my "proof" and everything fits perfectly now.

Atom
  • 4,620
  • 3
    The topological definition of "continuity at a point" you proposed is wrong. It only requires "For every neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$", no need to satisfy that $f^{-1}(V)$ is open. And you can easily prove this weaker definition is the correct one equivalent to the metric continuity. – Zerox Jun 16 '22 at 07:23
  • @Zerox what do you mean? What's your definition of neighborhood then? – JohannesPauling Jun 16 '22 at 07:26
  • 2
    A set $V$ is called a neighborhood of $x \in V$ iff $V$ contains an open set $U$ that contains $x$. – Zerox Jun 16 '22 at 07:27
  • I believe https://math.stackexchange.com/questions/2762135/equivalence-of-continuity-between-metric-and-topological-spaces may answer your question. – JohannesPauling Jun 16 '22 at 07:28
  • @Zerox I see; for me (following Munkres) a neighborhood $x\in U$ is an open set containing $x$. – JohannesPauling Jun 16 '22 at 07:30
  • Then the two definitions of "continuity at a point" you proposed are not equivalent. You can never deduce from the metric definition that $f^{-1}(B)$ is open if we only have the continuity at $x$. – Zerox Jun 16 '22 at 07:30
  • Where did you get the second definition from? – José Carlos Santos Jun 16 '22 at 07:38
  • @Zerox I think your definition implies mine, as I have proven in the "answer" below. But then, I might have made some error in the proof. – Atom Jun 16 '22 at 07:57
  • 1
    You get wrong in the proof. $f$ is not supposed to be continuous at $y$, so the neighborhood $U \ni y$ that is mapped into $V \ni f(y)$ may not exist. Remember in the definition only when $x$ is a continuous point could we find a neighborhood of $x$ mapped into a selected neighborhood of $f(x)$. – Zerox Jun 16 '22 at 08:01
  • 1
    It is a bit more esthetically pleasing to define continuity at a point by: $f: X\to Y$ is continuous at $x\in X$ if for every neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$. Here a neighborhood is understood in the sense of Bourbaki. – Moishe Kohan Jun 16 '22 at 09:17
  • IMO, rather than switching definition of neighborhood. I think, following is the conventional way to define continuity at a point in topological sense: $f:X\to Y$ is continuous at $x\in X$ iff $\forall V\in \mathcal{N}_{f(x)}$, $\exists U\in \mathcal{N}_x$ such that $f(U)\subseteq V$. Here is more equivalent definition of continuity. Here is proof of equivalent definition of continuity in metric space and topological sense. – user264745 Jun 16 '22 at 15:39

2 Answers2

3

A definition that avoids the use of the word "neighborhood" (so avoids a collision Munkres/Bourbaki).

$f:X\to Y$ is continuous at $x\in X$ if for every open set $U$ with $f(x)\in U$ an open set $V$ exists with $x\in V$ and $f(V)\subseteq U$.

Of course here you can replace $f(V)\subseteq U$ by the equivalent statement: $V\subseteq f^{-1}(U)$.

For $x=1$ and $U=(0,2)$ in your example e.g. $V=(0,2)$ does the job, and of course there are more choices for $V$.

drhab
  • 153,781
0

The function f IS NOT CONTINUOUS everywhere. It is continuous on (0,2) and then you get $f^{-1}$(0,2)=(0,2) which is open! If you are interested in one point x=1 then the definition is :Let f be a function defined from topological space X to topological space Y, then f is said to be continuous at a point x∈X if for every neighborhood V of f(x), there exists a neighborhood U of x, such that f(U)⊆V ! It says THERE EXISTS it does not say for all neighborhoods of a. So 1 is certainly a continuity point because if V=(0,2) THERE EXISTS a U=(1-ε,1+ε) such that f(U)$\subseteq $ V=(0,2). I cant make it clearer!