Let $X$ be a compact topological space and let $f : X \rightarrow \mathbb{R}$ be a function . The graph $f$ is the set $G = \{ (x, f(x) ) : x \in X \} \subseteq X \times \mathbb{R}$

My question is that Is the following statement is True/false ?

If $f$ is continuous, then $G$ is connected

My attempt : I think yes, because the continuous image of a connected set is connected.

  • $\begingroup$ you should at least assume that $X$ is connected. $\endgroup$
    – Thomas
    May 26 '19 at 7:24
  • $\begingroup$ @Thomas compact implies connected here already given that X be a compact topological space $\endgroup$
    – jasmine
    May 26 '19 at 7:25
  • 1
    $\begingroup$ nonsense: a finite metric space is compact but not connected. The Cantor set too. Or $[0,1] \cup \{2\}$ etc.. $\endgroup$ May 26 '19 at 7:26
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    $\begingroup$ @jasmine Compact most definitely does not imply connected, not even by a long shot. $\endgroup$
    – DonAntonio
    May 26 '19 at 7:26
  • $\begingroup$ @HennoBrandsma sir u r right i forget the example $\endgroup$
    – jasmine
    May 26 '19 at 7:27

False, and compactness is a red herring. We need $X$ to be connected.

It's clear that $G$ is homeomorhic to $X$ when $f$ is continuous (the first projection is the continuous inverse and when $f$ is continuous, so is the map $F: x \to (x, f(x))$ and $G=F[X]$).


In general, it is false. If, for instance, $X=\{0,1\}$, endowed with the discrete topology, then $X$ is compact, but $G$ is always disconnected.

Of course, the statement holds if $X$ is connected, even without assuming compactness.


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