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Now, there is probably more than hundreds of question asking what a neighborhood is, and trust me I've seen most of them. I understand the utility of neighborhoods is that some statemenets are easier to formulate with them eg. My favourite answer in regards to this topic has been this answer. I am also aware that there is a difference between american and non american textbooks (see) on how neighboorhood is defined. I take it to be defined as follows:

For a point $p$ of a topological space $(X,\tau)$ a neighboorhood $N$ is a subset of $(X,\tau)$ which contains a $p$ and also an open set $U_p$

And, I take open set to be defined as those elements in $\tau$.

Now, we got the conventions out of the way, let us move to the actual question.

A function $f:(X,\tau_X) \to (Y,\tau_Y)$ is said to be continous at a point $ a \in (X,\tau_X)$ , then for preimage of any neighborhood of $f(a)$ is a neighborhood.

Why couldn't we rather say preimage of any open set about $f(a)$ is open ? I think it may have something to do that continuity at a point not saying anything about continuity anywhere else see eg 1, but I can't make the issue precise.

Clemens Bartholdy
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  • Continuity is defined the way you suggest. In this case, we are considering functions that are not necessarily continuous, but are continuous at a single point. – John Douma Sep 05 '24 at 14:14
  • What about an example such as $f(x) = \begin{cases} x^2, & -1 \le x \le 1, \ 3, & x < -1 \lor x > 1 \end{cases}$. Then $(-1, 2)$ is an open set containing $f(0)$, but its preimage is $[-1, 1]$ which is not an open set -- though it is obviously a neighborhood of 0, as expected since $f$ is continuous at 0. Does that counterexample answer your question, or are you asking more abstractly why continuity at a point is defined the way it is? – Daniel Schepler Sep 05 '24 at 14:59
  • That is quite helpful. But, I'd also be interested for counter examples of same form which exist in general topological spaces since I have not found such a thing even after very aggresive searching through the site – Clemens Bartholdy Sep 05 '24 at 15:22

1 Answers1

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Your definition

$f$ is continuous at $x$ if the preimage of any open set containing $f (x)$ is open.

is stronger than the cited definition

$f$ is continuous at $x$ if the preimage of any neighborhood of $f (x)$ is a neighborhood (of $x$*).

Assume your definition. Let $W$ be a neighborhood of $f (x)$; then there is some open set $U \subseteq W$ that contains $f (x)$. The preimage $f^{-1} (U)$ is open by hypothesis and contains $x$. Obviously, $U \subseteq W \implies f^{-1} (U) \subseteq f^{-1} (W)$. Therefore, $f^{-1} (W)$ is a neighborhood of $x$.

The fact that your definition is not implied by the cited definition is shown by the good example in the comments.

Your definition is not really "local" in nature; it asks too much. Intuitively, a function should be continuous at a point $x$ if, no matter how small we define a neighborhood - WLOG, open set - $U$ about $f (x)$, the preimage of that neighborhood, $f^{-1} (U)$, is a neighborhood of $x$. There is no need for $f^{-1} (U)$ to be open, since, roughly speaking, we should not care about what happens at points different from $x$.

However, if it is the case that this property holds for all points $x \in X$, then of course the cited definition would also imply your definition. This is also easy to show. Assume the cited definition, and let $U$ be an open set containing $f (x)$. To show that $f^{-1} (U)$ is indeed open, let $y \in f^{-1} (U)$; this of course means that $f (y) \in U$: indeed, $U$ is a neighborhood of $f (y)$. Since $f$ is continuous at $y$, then by hypothesis, $f^{-1} (U)$ is a neighborhood of $y$, as desired.

* It obviously can't be just any neighborhood. A simple example is $f: \mathbb{R} \to \mathbb{R}$ where $f (x) = x$ if $x \ne 0$ and $f (0) = 1$; this function is not continuous at $x = 0$, but the preimage of any neighborhood of $f (0) = 1$ is a neighborhood of some point.

K. Jiang
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