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In the previous posts, I ask the existence of a continuous real-valued function on compact Hausdorff space which gives two distinct points in $X$ gives different values.

Compact Hausdorff space with distinct points $x,y$ implies existence of continuous real function $f$

Now I want to extend this more than two points. For examples

Let $X$ be a compact Hausodrff space and $x_i \in X$ be distints points in $X$. Then is there a continuous function $f:X \rightarrow \mathbb{R}$ such that $f(x_i) \neq f(x_j)$? [Let simply just define $f(x_i) =i$]

It seems since $X$ is normal, $n=2$ two points are okay. How about $n=3, 4,? $ or infinitely many points?

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  • $\begingroup$ What did you try? At least for finite case, it's still Uryshon lemma... $\endgroup$ Jul 20 at 8:04
  • $\begingroup$ @ArcticChar, How about the infinite case? Can we still use Uryshon lemma? $\endgroup$
    – phy_math
    Jul 20 at 8:08
  • $\begingroup$ Well, what do you think? Take a look at the statement of the lemma. $\endgroup$ Jul 20 at 8:10
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For a finite set of points, this follows from the Tietze extension theorem because compact Hausdorff spaces are normal, in a Hausdorff space, any finite set of points forms a closed subset, and all functions from a discrete set (finite set of points in a Hausdorff space) are continuous. (You can also use Urysohn's lemma and induction to get this)

That is, for your finite subset $A = \{x_1 \ldots, x_n \mid x_i \in X\}$, let $f \colon A \to \mathbb{R}$ be given by $f(x_i) = i$. Then the Tietze extension theorem says there exists a continuous function $\tilde{f} \colon X \to \mathbb{R}$ such that $\tilde{f}(x_i) = f(x_i)$.

For infinite (say countable) subsets, I believe the statement of the result breaks down in full generality because you can't guarantee the set $A$ is closed, though I can't think of a countable counterexample at the moment.

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  • $\begingroup$ In the countable case, $f(x_i) = i$ for all $i$ is not possible since $f$ has to be bounded. $\endgroup$ Jul 20 at 9:50
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    $\begingroup$ Yes, I explicitly said this argument only works for finite $A$. The countable case might be rather subtle, as it's not unreasonable to expect every countable subset of (say) $S^1$ can be separated by a real function. But I'm not sure. $\endgroup$
    – Dan Rust
    Jul 20 at 9:52
  • $\begingroup$ I mean, if the OP's question is: given a countably infinite subset $\{x_i\}$ of $X$, can we find $f$ such that $f(x_i) = i$? Then the answer is no, instead of not sure. $\endgroup$ Jul 20 at 9:55
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    $\begingroup$ Right, but I feel that parenthetical from the OP was just poorly thought through and not important. $\endgroup$
    – Dan Rust
    Jul 20 at 9:59
  • $\begingroup$ Ok, that's fair. To be honest, now that the OP has accepted another answer, I am not really sure what kind of question the OP has in mind. $\endgroup$ Jul 20 at 10:00
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If $X$ is compact, then every injective continuous map from $X$ into $\Bbb R$ is a homeomorphism from $X$ onto $f(X)$. So, if it turns out that $X$ is not homeomorphic to a subspace of $\Bbb R$ (such as when $X=S^1$), then the answer is negative.

In the finite case, it is true, by Urysohn's lemma, as you were told in the comments.

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    $\begingroup$ "every continuous map from $X$ into $\mathbb{R}$ is a homeomorphism" you are confusing me $\endgroup$
    – Didier
    Jul 20 at 8:38
  • $\begingroup$ I guess you want injectivity of the continuous map? Even with that, it's not clear how this answers the question: that $f$ is injective when restricted to the points $\{x_i\}$ does not mean that $f$ itself is injective. $\endgroup$ Jul 20 at 8:40
  • $\begingroup$ Also, of course one cannot have $f(x_i) = i$ by compactness, but there are definitely cases where one can find continuous map so that $\{f(x_i)\}$ are all distinct. $\endgroup$ Jul 20 at 8:43
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    $\begingroup$ The op said nothing about the function needing to be injective... $\endgroup$
    – Dan Rust
    Jul 20 at 8:54
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    $\begingroup$ That is, the op is asking if the set $\{x_1, \ldots, x_n\mid x_i\in X\}$ can be separated by a continuous function as long as X is compact Hausdorff $\endgroup$
    – Dan Rust
    Jul 20 at 9:06

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