I was trying to prove the equivalence between the epsilon delta definition and open ball definition of continuity, and there have already been quite a few discussions such as here already. I realised that in proving that "epsilon delta $\Rightarrow$ open ball" we try to show that every point in the preimage of an open ball in the domain is an interior point, which requires the global $\epsilon-\delta$ continuity of a function. This seems to suggest to me that open ball continuity is stronger, as in these two examples:

1) $f(x) :=\begin{cases} 0 & x \text{ is rational}\\ x & x\text{ is irrational} \end{cases}$

Here f(x) is continuous only at $x = 0$. However, the preimage of any open ball containing zero is never open.

2) $ f(x):=\begin{cases} 1 & x<-1 \\ 0 & -1\leq x \leq 1 \\ 1 & x>1 \end{cases} $

Here $f(x)$ is continuous at zero, yet there exist an open ball $(-1/2,1/2)$ whose preimage is not open.

If continuity on functions only 'makes sense' for global continuity, why do we then still talk about continuity at a point in a topological space (i.e. a function is continuous at $x$ if every neighbourhood of $x$ pulls back to open sets) ?

  • $\begingroup$ @PeterFranek Thanks, I have corrected it. $\endgroup$ – Zhanfeng Lim Feb 14 '17 at 14:00

As you find in this post, $f(x)$ being continuous in a certain point $x\in X$ is topologically defined in the following way: If $V$ is an open neighborhood of $f(x)$, then the preimage of $V$ contains an open neighborhood $U$ of $x$.

When thinking about $V$ being of size $\varepsilon$ and $U$ being of size $\delta$, this resembles the $\varepsilon$-$\delta$-definition in an intuitive way.

  • $\begingroup$ How does this work out for the first example? The preimage of a open set containing 0 contains $\{0\}\cup \{Irrational numbers\}$, doesn't it contain no open neighbourhood? $\endgroup$ – Zhanfeng Lim Feb 14 '17 at 15:49
  • $\begingroup$ The preimage of an open set $V$ around 0 will contain all rationals (as they are mapped to zero) and some irrationals around 0 (irrationals with an absolute value small enough to be mapped to $V$). So some open set around 0 gets completely mappen to $V$. This is your $U$. So your difficulty apparantly arose when computing the preimage. $\endgroup$ – M. Winter Feb 14 '17 at 16:01
  • $\begingroup$ What an embarassing mistake, thanks for pointing that out! $\endgroup$ – Zhanfeng Lim Feb 14 '17 at 16:11

You can define a function $f: X\to Y$ to be continuous at a point $x$ iff for every open set $N$ containing $x$ there is an open set $V\subseteq Y$ containing $f(x)$ such that $f^{-1}(V)\subseteq N$. In words, preimage of a neighborhood is inside a neighborhood. So yes, you can talk about "continuity at a point" in the topological category, if this is what you ask for.

This definition is of course compatible with all other definitions $(\epsilon-\delta)$ and indeed, continuity at a point is a topological property.


You can consider this a long comment, if it doesn't suit your needs as an answer.

It is my view (as a student of analysis, but not an expert) that analysis is largely about dealing with infinitely large and small things. A lot of times, we get away with just using our intuition about these things, because not much can or does go wrong - this is how we do pretty much all of one variable and even several variable calculus. But a phenomenon that happens quite often is that two or more processes which we can intuitively understand can clash in ways that we may not understand anymore. A natural thing to hope for is that when two such concepts meet, they are interchangeable, in the sense that our intuition about them is preserved, and the order in which we turn our attention to them doesn't matter.

This is the case for continuity at a point. Another notion of continuity which you will probably have seen is that $f$ is continuous at $x$ if, for every sequence $x_n \to x$, we will have $f(x_n) \to f(x)$. In other words, $f$ respects sequences. We know how functions should behave, of course, assuming we know the function, and we know how the sequence behaves - we have prescribed it's behavior to converge to $x$. But do these two concepts play nice together? Another way of saying this is $\textit{can we exchange the limit with the function?}$. The answer is yes, provided $f$ is continuous at the limit point. It doesn't matter if we first take the limit, and then apply $f$ to the limit point, or if we map the whole sequence, and take the limit.

There are many theorems in analysis that amount to 'exchange of limit' theorems, and in some sense, these are the bulk of elementary analysis, like you would find in Rudin.


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