# Showing that a set is meager

Consider $X =C([-1,1])$ with the usual norm $\lVert f \rVert_{\infty} = \sup_{t\in [-1,1]} |f(t)|.$ Let \begin{align*} A_+ &= \{ f \in X: f(t) = f(-t) \quad \forall t \in[-1,1] \}\\ A_- &= \{ f \in X: f(t) = -f(-t) \quad \forall t\in [-1,1] \}. \end{align*} Show that $A_+$ and $A_{-}$ are meager.

I'm uncertain on how to proceed with this.

• You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed? – Mathematician 42 Dec 9 '17 at 21:03
• As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior. – Martin Sleziak Dec 10 '17 at 11:20

For each of $$A_{+}$$ and $$A_{-}$$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $$\varepsilon$$-ball around a point $$f$$ looks like. It is the set of all continuous functions that differ from $$f$$ by less than $$\varepsilon$$ at any point in $$[0,1]$$. That is, the set of of all continuous functions that lie within a $$\varepsilon$$ “tube” around $$f$$.
To see that the $$A_{+}$$ is closed, we show its complement $$U$$, is open. If $$g \in U$$, then $$\exists T \in [0,1]$$ such that $$g(-T) \neq g(T)$$. Define $$\delta = g(T)-g(-T)$$. Now for any continuous function $$h$$ in the $$\frac{\delta}{3}$$ ball around $$g$$, it’s clear that $$h(T) \neq h(-T)$$. In other words, the entire $$\frac{\delta}{3}$$ ball around $$g$$ lies within $$U$$. So $$U$$ is open and therefore, $$A_{+}$$ is closed.
To show its interior is empty, pick any $$f\in A_{+}$$ and consider an $$\varepsilon$$-ball $$B$$ around $$f$$. Since this is an $$\varepsilon$$-tube around $$f$$, it is easy to see that there are continuous functions $$s\in B$$ such that for some $$T, s(T) \neq s(-T)$$. In other words, $$A_{+}$$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $$A_{-}$$.
• Seems easier to me to use sequences to show $A+$ is closed. – zhw. Dec 9 '17 at 23:47