My question occurs when I see the problem that show that every continuous function is borel measurable. I know that for topological space $(X,\tau)$, we define that "open sets" means sets in topology. However, it is somewhat weird for me in this example;

Let $\tau$ be trivial (indiscrete) topology, which means that $\tau = \{X, \emptyset \}$. Let $X = \mathbb{R}$. Then only open set in this topological space is $\mathbb{R}, \emptyset$. However, we already know that any interval $(a,b)$ where $a,b \in \mathbb{R}$ is open!

So let's see the example of continuous function $f : (X,\tau) \to (X, 2^{X})$. In this case, does it really matter that any preimage of open sets in $2^{X}$ except $\emptyset$ is $X$ itself? I think this is intuitively confusing and somewhat contradiction occurs.

Am I thinking wrong or missing something on definition of topology?

  • $\begingroup$ Intervals of this form are open in the usual (i.e. Euclidean) topology on $\mathbb{R}$, but not others. $\endgroup$ – user61527 Jan 23 '14 at 6:28
  • $\begingroup$ @T.Bongers So you mean that in this trivial topology on $\mathbb{R}$, all interval is not open by axiom? $\endgroup$ – user122655 Jan 23 '14 at 6:30

Open intervals in $\mathbb{R}$ are not always open. What you are thinking about is what is called the usual topology on $\mathbb{R}$. This topology is defined to be the one generated by open balls, i.e. taking $B_\delta(x)=\{a\in\mathbb{R}|d(x,a)<\delta\}$ and then carrying out intersections and unions to discover what other sets must be open. If we have a different topology, then different sets are open.

The example of the topology you mention of just having $\{\},X$ be open is called the indiscrete topology. The discrete topology is when every set is open.

  • $\begingroup$ Oh I see. Thank you! And for clarity of question, I edit my question now. $\endgroup$ – user122655 Jan 23 '14 at 6:34
  • $\begingroup$ Can I ask one more question? then if $f$ is continuous on this example, then for any open set $A$ in $2^{X}$, $f^{-1}(A) = X$? $\endgroup$ – user122655 Jan 23 '14 at 6:37
  • $\begingroup$ Yes. In topology, that is the definition of continuous. The analysis definition of continuous actually comes from that one. Note that how I defined a ball can be used in any metric space. Thus said construction for open sets does too. In that topology, continuous means what it intuitively does in terms of calculus and functions one is used to seeing. In this example, $f(x)=x$ is not continuous (probably, assuming $2^X$ means the indiscreet topology since it's the power set) $\endgroup$ – Stella Biderman Jan 23 '14 at 6:40
  • $\begingroup$ Wow, Thank you! Your kind explanation and example makes me understand! $\endgroup$ – user122655 Jan 23 '14 at 6:49

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