Consider $X=C[0,1]$ with its 'sup-norm' topology. Let $$S=\Bigg\{ f \in X: \int_0^1f(x)\;dx \neq 0\Bigg\}$$ How to prove $S$ is dense in $X$ ?

My try: I know this result " If a subset $A$ is nowhere dense in $X$, then $X \setminus A$ is dense in $X$"

The equivalent statement is " If $X \setminus A$ is NOT dense in $X$, then $A$ is NOT nowhere dense in $X$.

Here, $X \setminus S$ is a proper closed set, so it is not dense in $X$ and hence by above equivalent statement, $S$ is dense in $X$.

Is this right? Any help?


1 Answer 1


You can prove this directly. For any $f\notin X$, it is easy to see that $f+\epsilon\in S$ for any $\epsilon>0$.

  • $\begingroup$ I don't understand what you are saying! I Know this definition: $S$ is dense in $X$ if every $f \in X$ is either belong to $S$ or a limit point of $S$ $\endgroup$
    – user444830
    Sep 11, 2018 at 2:39
  • $\begingroup$ $S$ being dense it means that any open ball $B(f, \epsilon)$ in the sup norm must contain some element of $S$. $\endgroup$
    – Eduardo
    Sep 11, 2018 at 2:45
  • $\begingroup$ @LDM You can choose a sequence $f+\frac1n\to f$. $\endgroup$ Sep 11, 2018 at 3:15
  • $\begingroup$ Ok! Whether my argument is right or wrong? $\endgroup$
    – user444830
    Sep 11, 2018 at 3:28
  • $\begingroup$ @EclipseSun: I think, you start with "For any $f \notin S$, it is easy ....." I don't know why you pick $f \notin X$ $\endgroup$
    – user444830
    Sep 11, 2018 at 5:50

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