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Let $F = \{ f \in C[-2,2] \mid f(0)+f(1)=3\}$. Does $F$ contain an open subset of $C[-2,2]$?

I'm drawing blank with this problem. I couldn't figure out any clues on whether it should or should not contain an open subset of $C[-2,2]$. If it did, then for $O \subset F$ open and for any $f \in O \subset F$ there should be an open ball at $f$ such that $B(f,r) \subset O \subset F$.

I don't find any contradictive results if I consider some $g \in B(f,r)$. The conclusions that can be drawn from here is that $\|f-g\|_\infty < r$. One thing is that if I could show that $g$ does not satisfy $g(0)+g(1) =3$, then as $f$ and $O$ are both arbitary I could conclude that $O$ cannot be open, but it would look like $g(0)+g(1) =3$ is true always if $g \in B(f,r)$.

Can I have some ideas what to consider in this problem?

Walker
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    No it does not for $F$ is an affine space (a translation of a vector space $V={f: f(0)+f(1)=3}$. Not proper linear subspace of a linear space has interior. – Mittens Oct 19 '22 at 20:42
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    You can show that the complement of $F$ is a dense set, and so intersects every open set. Given $f\in F$, can you find a sequence of functions not in $F$ which converges to $f$? – Mark Oct 19 '22 at 20:43
  • @OliverDíaz What do you exactly mean by a translation of a vector space $V={f: f(0)+f(1)=3}$? Is $V$ not equal to $F$? – Walker Oct 19 '22 at 20:53
  • @SleepWalker: I mean to say $V={f: f(0)+f(3)=0}$. – Mittens Oct 19 '22 at 20:57
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    If $f\in F$ then $f+{1\over n}\notin F$ for any $n.$ Hence $f$ does not belong to the interior of $F.$ – Ryszard Szwarc Oct 20 '22 at 05:14

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