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I am trying to understand whether a subbase for compact-open topology is a base for the topology?

So, let $X,Y$ be topological space, and $C(X,Y)$ the space of continuous functions from $X$ to $Y$.

For a compact $K\subset X$ and open $V\subset Y$, let $M(K,V)$ denote the collection of all those continuous function $f:X\rightarrow Y$ with $f(K)\subset V$.

The compact open topology on $C(X,Y)$ has subbase $\{M(K,V)\}_{K,V}$ with $K\subset X$ compact and $V\subset Y$ open.

For compact sets $K_1, K_2$ in $X$ and open $V_1,V_2$ in $Y$, I do not understand whether $M(K_1,V_1)\cap M(K_2,V_2)$ is always $M(K,V)$ for some compact $K$ in $X$ and open $V$ in $Y$. Any example for truth/failure of this fact?

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    $\begingroup$ Try $X=\{a,b\}$ discrete, $K_1=\{a\}$, $K_2=\{b\}$. $\endgroup$ – Lord Shark the Unknown Dec 28 '17 at 13:15
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Let $X=Y=\mathbb R$ with the usual topology. Then, there exist no $K\subseteq \mathbb R$ and $V\subseteq\mathbb R$ such that

  • $K$ is compact,
  • $V$ is open, and
  • $M(\{0,1\},(0,1))\cap M(\{1,2\},(0,2))=M(K,V)$.

To see this, suppose, for the sake of contradiction, that such $K$ and $V$ exist. For brevity, let \begin{align*} M_1\equiv&\;M(\{0,1\},(0,1)),\\ M_2\equiv&\;M(\{1,2\},(0,2)) \end{align*} in what follows.

Claim 1$\phantom{--}$ $V\neq \mathbb R$.

Proof$\phantom{--}$ If $V=\mathbb R$, then $M(K,V)$ contains all real-to-real continuous functions, but $M_1\cap M_2$ clearly does not contain all real-to-real continuous functions. $\phantom{--}\blacksquare$

Claim 2$\phantom{--}$ $V\neq \varnothing$.

Proof$\phantom{--}$ If $V=\varnothing$, then the only way that $M(K,V)$ can be non-empty (which it must be, given that $M_1\cap M_2$ is obviously non-empty) is by having $K=\varnothing$. But then $M(K,V)$ contains all real-to-real continuous functions, which is, again, impossible. $\phantom{--}\blacksquare$

Claim 3$\phantom{--}$ $K=\{0,1,2\}$.

Proof$\phantom{--}$ If $x\in K$ but $x\notin\{0,1,2\}$, then construct a continuous function (through, say, linear interpolation) $f:\mathbb R\to\mathbb R$ such that $f(0)=f(1)=f(2)=1/2$ and $f(x)=v$ for some arbitrary $v\in\mathbb R\setminus V$ (see Claim 1). Then, $f\in M_1\cap M_2$, but $f\notin M(K,V)$. Therefore, one must have $K\subseteq\{0,1,2\}$. Conversely, if $x\in\{0,1,2\}$ but $x\notin K$, then pick any $v\in V$ (which is possible by Claim 2) and construct a continuous function $f:\mathbb R\to\mathbb R$ such that $f(x)=3$ and $f(y)=v$ for $y\in\{0,1,2\}\setminus\{x\}$. It is easy to check that $f\in M(K,V)$ (remember it has been already proven that $K\subseteq\{0,1,2\}$) but $f\notin M_1\cap M_2$. Therefore, one must have $\{0,1,2\}\subseteq K$. $\phantom{--}\blacksquare$

Claim 4$\phantom{--}$ $(0,2)\subseteq V$.

Proof$\phantom{--}$ If not, then take any $v\in(0,2)$ such that $v\notin V$. Construct a continuous function $f:\mathbb R\to\mathbb R$ such that $f(0)=f(1)=1/2$ and $f(2)=v$. But then $f\in M_1\cap M_2$, yet $f\notin M(K,V)$, given also that $K=\{0,1,2\}$ by Claim 3. $\phantom{--}\blacksquare$

To obtain a final contradiction, consider the constant function $f\equiv 1$. By Claim 4, $f\in M(K,V)$, but $f\notin M_1\cap M_2$.

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Expanding upon the comment by Lord Shark, we construct a counterexample. Take any space $Y$ which contains two disjoint open sets $V_1$ and $V_2$. Now $M(K_i,V_i)$ is the set of functions which map $a$/$b$ into $V_1$/$V_2$. The intersection $F = M(K_1,V_1)\cap M(K_2,V_2)$ is the set of all functions which map $a$ into $V_1$ and $b$ into $V_2$.

I claim, however, that this is not $M(K,V)$ for any sets $K$ and $V$. To see this, note that $f\in F$ is equivalent to both a restriction both on $f(a)$ and $f(b)$. Thus if $F = M(K,V)$ we must have $K=X$. Also, since $F$ contains all functions mapping $a$ into $V_1$ and $b$ into $V_2$, we must have $V_1\cup V_2\subset V$ (since the image of $K=X$ can be any two-point subset with one point in $V_1$ and one point in $V_2$). But then $M(K,V)$ contains some function $g$ mapping both $a$ and $b$ into $V_1$, and since $V_1$ and $V_2$ are disjoint we can not have $g\in F$.

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  • $\begingroup$ Maybe I'm not fully understanding what you mean, but you could word your argument differently in claiming that $K=X$; Every subset of $X$ is compact, as there are only finitely many open sets. But clearly, for example, the choice $K=\{a\}$ is not valid as there are some functions such that $f(a)\in V_1$ while $f(b)$ can be anything. Same for $\{b\}$. So we are left with $X$ $\endgroup$ – user160738 Dec 28 '17 at 15:21
  • $\begingroup$ Yes, the argument is that the set "$K$" in this notation is the only one on whose image we place an explicit condition. But when $X$ has only two points, and the image of each is restricted, this set on which we place the restriction must be all of $X$. $\endgroup$ – mcwiggler Dec 29 '17 at 0:19

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