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Nowhere-dense sets are small in certain sense; for example, the Cantor set and \(\mathbb{Z}\) both are nowhere-dense in \(\mathbb{R}\). Also every single element set \(\{x\}\) is nowhere-dense in \(\mathbb{R}\). Note that a set \(S\) is nowhere-dense in \(\mathbb{R}\) if it is dense in no (open) interval at all, or equivalently every interval \(I\subset\mathbb{R}\) contains another interval \(I^{*}\) such that \(I^{*}\cap S=\emptyset\) (check the equivalence).
**Theorem 65** (Baire’s Theorem for \(\mathbb{R}\)).: _The set of real numbers \(\mathbb{R}\) cannot be written as the countable union of nowhere-dense sets._
Proof
: Suppose to the contrary that \(\mathbb{R}=\bigcup_{n=1}^{\infty}A_{n}\) where every set \(A_{n}\) is nowhere-dense in \(\mathbb{R}\). Let \(I_{0}\subset\mathbb{R}\) be a closed interval of radius \(1\). Since \(A_{1}\) is nowhere-dense in \(\mathbb{R}\), it is nowhere-dense in \(I_{0}\); thus there is a closed interval \(I_{1}\) of radius less than \(1/2\) such that \(I_{1}\subset I_{0}\) and \(I_{1}\cap A_{1}=\emptyset\). Now observe that \(A_{2}\) is nowhere-dense in \(I_{1}\), thus there is a closed interval \(I_{2}\) of radius less than \(1/3\) such that \(I_{2}\subset I_{1}\) and \(I_{2}\cap A_{2}=\emptyset\). Continuing in this fashion we get a sequence of closed intervals \(\{I_{n}\}\) with radii converging to zero such that
\[I_{n}\cap A_{n}=\emptyset\quad\text{for}\quad n=1,2,\ldots\]
By the Nested Intervals Theorem the intersection \(\bigcap_{n=1}^{\infty}I_{n}\) contains a point \(x\in\mathbb{R}\). However, by the above construction \(x\notin\bigcup_{n=1}^{\infty}A_{n}\), so it follows that \(\mathbb{R}\) cannot be written as \(\bigcup_{n=1}^{\infty}A_{n}\) where each \(A_{n}\) is nowhere-dense in \(\mathbb{R}\).
**Corollary 12**.: \(\mathbb{Q}\) _cannot be written as the countable intersection of open sets._
Proof
: Observe that a dense \(G_{\delta}\) subset of \(\mathbb{R}\) must be also uncountable.
**Remark 36**.: Notice how very different \(\mathbb{Q}\) and \(\mathbb{I}=\mathbb{R}\setminus\mathbb{Q}\) are. Irrationals are quite “thick” while rationals are “sparse.” As we have stated before, replacing \(\mathbb{R}\) with an arbitrary complete metric space \(M\), one can prove Baire’s Theorem in this more general setting which is also called Baire’s Category Theorem. The word _category_ is there because it creates two categories of size for subsets of a metric space. A set qualifies to be “first category” if it can be written as a countable union of nowhere-dense sets; sometimes the word “meager” is used instead of first category. These are thin subsets of a metric space. Baire’s Category Theorem asserts that a complete metric space is not the union of a countable collection of nowhere-dense sets. Thus, if the given metric space is complete, it is necessarily of the “second category,” i.e., any complete metric space must be too large to be the countable union of nowhere-dense sets. Baire’s Category Theorem plays an important role in many parts of analysis, in particular, in measure theory and the study of bounded linear operators on Banach spaces (a Banach Space is a complete normed linear space). In the following we state a very interesting and unexpected consequence of Baire’s Theorem.
**Theorem 66** (An application of Baire’s Theorem).: _Given the complete metric space \((C[0,1],d)\) where \(d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|\), let \(S\subset C[0,1]\) be defined as_
\[S=\{f\in C[0,1]:\ f^{\prime}(x)\text{ exists for some }x\in[0,1]\}.\]
_Then the set \(S\) is of the first category in \(C[0,1]\)._
The above theorem asserts that most continuous functions do not have derivatives at any point, or continuous functions that are differentiable at even one point in \([0,1]\) form a “sparse” subset. This is not intuitive, but an unexpected consequence of Baire’s Theorem: most continuous functions are nowhere differentiable! For a construction of such a function we refer the reader to, e.g., [1], pp. 146-148.
### Inner Product Spaces
The concept of a vector space generalizes what we already know about addition and scalar multiplication on \(\mathbb{R}\) or \(\mathbb{R}^{2}\). However, if we want to talk about the concept of _length_ or _angle_ in a vector space, we need to study inner product spaces. To motivate the concept of inner product spaces, consider \(x\in\mathbb{R}^{2}\) with initial point at the origin.
Then the vector \(x=(x_{1},x_{2})\) has a _length_ which is called the _norm_ of \(x\), denoted by \(||x||\) (Figure 2.6) and defined as
\[||x||=\sqrt{x_{1}^{2}+x_{2}^{2}}.\]
An obvious generalization of the norm of \(x\in\mathbb{R}^{n}\) is:
\[||x||=\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}.\]
To utilize linearity in our discussion we introduce the dot product.
**Definition 56** (Dot Product).: The dot product of two vectors \(x,y\in\mathbb{R}^{n}\) where \(x=(x_{1},x_{2},\ldots,x_{n})\) and \(y=(y_{1},y_{1},\ldots,y_{n})\) is defined as
\[x\cdot y=x_{1}y_{1}+\cdots+x_{n}y_{n}=\sum_{i=1}^{n}x_{i}y_{i}.\]
This number relates to the norm by
\[||x||^{2}=x\cdot x.\]
Note that \(x\cdot x\geq 0\) for all \(x\in\mathbb{R}^{n}\) with equality in the case \(x=0\). Furthermore, \(x\cdot y=y\cdot x\) for all \(x,y\in\mathbb{R}^{n}\) and if we keep \(y\in\mathbb{R}^{n}\) fixed, we can define the linear map
\[\Phi:\mathbb{R}^{n}\to\mathbb{R}\quad\text{ by }\quad\Phi(x)=x\cdot y.\]
An _inner product_ is a generalization of a _dot product_. Thus the question we try to answer is if we consider the above properties of a dot product and define an inner product by abstracting these concepts, will we get all the properties of inner product spaces? It turns out that the answer is yes for **real** vector spaces. But if we want to define inner product for real as well as **complex** vector spaces we need to recall some concepts about complex numbers. Let \(z=x+iy\in\mathbb{C}\) where \(x,y\in\mathbb{R}\). The absolute value \(|z|\) of \(z\) and the complex conjugate \(\overline{z}\) of a \(z\) are defined as:
\[|z|=\sqrt{x^{2}+y^{2}},\quad\overline{z}=x-iy.\]
Figure 2.6: The norm of \(x\)
Note that
\[z\overline{z}=(x+iy)(x-iy)=x^{2}+y^{2}=|z|^{2}.\]
Now, we are ready to define a complex norm.
**Definition 57** (Complex Norm).: For \(z\in\mathbb{C}^{n}\) where \(z=(z_{1},z_{2},\ldots,z_{n})\), then the norm of \(z\) is defined as
\[||z||=\sqrt{|z_{1}|^{2}+|z_{2}|^{2}+\cdots+|z_{n}|^{2}},\]
where we need absolute values because \(||z||\) is a non-negative number.
Note that
\[||z||^{2}=|z_{1}|^{2}+|z_{2}|^{2}+\cdots+|z_{n}|^{2}=z_{1}\overline{z_{1}}+z_{2 }\overline{z_{2}}+\cdots+z_{n}\overline{z_{n}}.\]
So we can define an inner product of \(z,w\in\mathbb{C}^{n}\).
**Definition 58** (Complex Inner Product).: For \(z,w\in\mathbb{C}^{n}\), we define the complex inner product as
\[z\cdot w=z_{1}\overline{w_{1}}+z_{2}\overline{w_{2}}+\cdots+z_{n}\overline{w_ {n}}.\]
**Example 65**.: Let \(z=(i,1+i,3)\) and \(w=(1-i,2,4i)\). Let us find \(||z||\), \(z\cdot w\) and \(d(z,w)\).
First,
\[||z||^{2} = z_{1}\overline{z_{1}}+z_{2}\overline{z_{2}}+z_{3}\overline{z_{ 3}}=(i)(-i)+(1+i)(1-i)+3\cdot 3\] \[= -i^{2}+1-i+i-i^{2}+9=1+1+1+9=12,\]
which implies \(||z||=\sqrt{12}.\) Then
\[z\cdot w = i(1+i)+(1+i)(2)+3(-4i)=i+i^{2}+2+2i-12i\] \[= 1-9i.\]
We know that \(d(z,w)=||z-w||\), so we will first determine \(z-w\).
\[z-w=(i,1+i,3)-(1-i,2,4i)=(-1+2i,-1+i,3-4i).\]
So we have
\[||z-w||^{2} = |-1+2i|^{2}+|-1+i|^{2}+|3-4i|^{2}\] \[= [(-1)^{2}+(2)^{2}]+[(-1)^{2}+(1)^{2}]+[(3)^{2}+(-4)^{2}]\] \[= 5+2+25=32,\]
which implies \(||z-w||=\sqrt{32}=4\sqrt{2}\).
**Definition 59** (Inner Product).: An inner product on a real or complex vector space \(V\) is a function
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