0
$\begingroup$

Question:

What topology must $X$ have if every real-valued function defined on $X$ is continuous?

Solution:

$X$ must have the discrete topology where every subset is open.

To show this it suffices to show points in $X$ are open.

Let $x \in X$. Define $f : X \rightarrow \mathbb{R}$ by

$f(x) = 0$ and $f(y) = 1 \:\:\forall\: y \:\neq \:x$.

Then $f^{−1} ((\frac{-1}{2}, \frac{1}{2})) = \{x\}$.

Thus if $f$ is continuous, since $(\frac{-1}{2}, \frac{1}{2})$ is open, $\{x\}$ is open.


Can anyone possibly explain the logic and reasons behind this solution to someone very new to topology?

$\endgroup$
  • 1
    $\begingroup$ That's a pretty unclear question. What in particular do you want explained? You're looking at a bump function on $X$, which is $1$ on $x$ and $0$ everywhere else. If this is continuous, $\{x\}$ has to be open. $\endgroup$ – Balarka Sen Jan 23 '16 at 19:51
  • $\begingroup$ Well i dont understand why it must have the discrete topology. I understand that it suffices to show that points in X are open to show that X have the discrete topology. Then they construct a function. I dont understand why its enough to use this particular function or what is happening on the fifth row of the solution. I understand tho that if f is continous its inverse takes open intervals to open intervals. $\endgroup$ – JKnecht Jan 23 '16 at 20:11
  • $\begingroup$ I started with topology a couple of days ago so its still pretty confusing to me... $\endgroup$ – JKnecht Jan 23 '16 at 20:13
  • 1
    $\begingroup$ They're saying every real valued function on $X$ is continuous. So in particular this "bump" function $f$ must be continuous too. But by definition, preimage of open sets by continuous functions is open. Thus, $f^{-1}(-1/2, 1/2)$ needs to be open, as $(-1/2, 1/2)$ is open. However, the only value $f$ takes in $(-1/2, 1/2)$ is $0$ and only when $f$ is applied to $x$, so $f^{-1}(-1/2, 1/2) = \{x\}$. So $\{x\}$ needs to be open. $\endgroup$ – Balarka Sen Jan 23 '16 at 20:16
  • $\begingroup$ @Balarka Sen...Yes now i am starting to understand. And i even understand why they can say that it must have the discrete topology. Thanks! $\endgroup$ – JKnecht Jan 23 '16 at 20:33
1
$\begingroup$

The codomain being $\mathbb{R}$ is a little misleading...all you need to know is that $\mathbb{R}$ contains a copy of $\{0,1\}$ with the discrete topology. (Note that the subspace $\{0,1\} \subset \mathbb{R}$ is such that the subspace topology is the discrete topology).

Any map $f:X \to \{0,1\}$ separates $X$ into two disjoint subsets--the preimage of $0$ and the preimage of $1$. Intuitively, because $0$ and $1$ are not "close together" in the codomain, this function is "breaking" $X$ into two pieces, one of which it assigns to $0$ and the other to $1$. If $f$ is surjective, these two pieces are nonempty.

In order for $f$ to be continuous (i.e. in order for $f^{-1}(\{1\})$ and $f^{-1}(\{0\})$ to be open), the pieces had to be already "broken apart" to begin with (since continuity means $f$ can't break anything). So if any such function is continuous, that means that anytime I break $X$ into two pieces $X_1$ and $X_2$, the sets $X_1$ and $X_2$ were already different connected components--in other words, they are both closed and open.

Since any subset $X_1 \subset X$ is open, $X$ has the discrete topology.

$\endgroup$
  • $\begingroup$ This is little over my head atm, but hopefully i will understand it in a couple of days. Thx. Anyway i get it now after reading Balarka Sen's comment. $\endgroup$ – JKnecht Jan 23 '16 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.