[locally-compact-blog]

Every metric is continuous

This fact struck me as very not obvious and super interesting. I thought I’d provide a proof of it here; this is problem 3a of section 20 in James R. Munkres’ Topology, 2nd. Edition.

Proof: Let $X$ be any metric space with some metric $d:X\times X\rightarrow\mathbb R$, and let $U$ be open in $\mathbb R$. Consider any point $(x,y)\in d^{-1}(U)$, and select a basis element $(\alpha, \beta)$ of $\mathbb R$ for which $d(x,y)\in (\alpha, \beta)\subset U$. Pick a number $\epsilon > 0$ such that  $d(x,y)\in (d(x,y)-\epsilon, d(x,y)+\epsilon)\subset (\alpha, \beta)$ (convince yourself that that’s reasonable if you have to--it seems to come in handy in a lot of these kinds of arguments). Now, consider the following sets: $$A=B_d(x, \frac{\epsilon}{2})$$ and $$B= B_d(y, \frac{\epsilon}{2}).$$ Since both $A$ and $B$ are open in $X$ due to the fact that $x$ and $y$ both came from $X$, $A\times B$ belong to a basis for $X\times X$; further, observe that $(x,y)$ itself belongs to $A\times B$. Let $(x_0, y_0)\in A\times B$. On the one hand, we have:

$$d(x,x_0)+d(y,y_0) < \epsilon$$ which implies

$$d(x_0,x)+d(x,y)+d(y,y_0) < d(x,y)+\epsilon$$

and further, by the triangle inequality and symmetry of $d$, that

$$d(x_0,y_0) < d(x,y)+\epsilon < \beta.$$

On the other hand, we have: $$d(x_0, y_0)+\epsilon > d(x_0, y_0)+d(x,x_0)+d(y,y_0)$$ and so

$$ d(x_0, y_0)+\epsilon > d(x,y)$$ and finally

$$d(x_0,y_0) > d(x,y)-\epsilon > \alpha.$$

Therefore $d(x_0, y_0)\in (\alpha, \beta)\subset U$, which means that $(x_0, y_0)\in d^{-1}(U)$, and so $(x,y)\in A\times B\subset d^{-1}(U)$ meaning that $d$ is continuous.                                                                                           ∎

[locally-compact-blog]

Simple proof that the space of complex valued polynomials is infinite dimensional

Let p = $(p_1,p_2,\ldots,p_n)$ be any finite list of polynomials in $p(\mathbb C)$ (the space of complex valued polynomials). Because this list is finite, one of the vectors in it has a degree greater than or equal to any other vector in the list. Let’s say that polynomial is of degree $n$.

Now, since $p(\mathbb C)$ contains every complex valued polynomial, in particular $p(z)=z^{n+1}$ is in $p(\mathbb C)$ for any given $n$. Notice that the set $span$(p) cannot contain such $z^{n+1}$, since no linear combination of polynomials of degree $n$ can yield one of degree $n+1$*. 

This tells us that there is at least one polynomial in $p(\mathbb C)$ that is not in $span$(p), which in turn tells us that $p(\mathbb C)$ is not a subset of  $span$(p). Since two sets are equal if and only if they are subsets of each other, it must be that $span$(p) $\neq$ $p(\mathbb C)$. Since p was arbitrary, we may conclude that there is no finite list of vectors in $p(\mathbb C)$ whose span yields $p(\mathbb C)$. 

However, a vector space is finite dimensional if and only if there does exist such a list. Therefore $p(\mathbb C)$ is not finite dimensional, as was to be shown.

* To see this, suppose there was such a linear combination. Equate the linear combination to $z^{n+1}$ and take n+1 derivatives. This yields a contradiction.

[theonion]

Anybody who knows me will tell you the same thing: I get what I want. Whether it’s food, being held, my binky, you name it—if I decide I’d like it, you damn well better believe I don’t rest until I get it, one way or another. And from the very second I saw those blue and red detergent pods come out of that shopping bag last week, I knew immediately that, come hell or high water, I would eat one of those things.

So with God as my witness, I swear to you: I’m going to find that container of multicolored pods, I’m going to take one out, I’m going to shove it in my mouth, and I’m going to chew it up and swallow it down, and nothing and no one is going to stand in my way.

More.

[locally-compact-blog]

Definition of a vector space

I’m going to be working through Sheldon Axler’s Linear Algebra Done Right for the next little while, so I believe it will be useful to document my understanding of the proofs and definitions here. 

Let $V$ be any set, and let $v_i\in V$ for any given natural number $i$; similarly, let $c_i\in\mathbb C$ for any given natural number $i$. Define functions $+$ and $\cdot$ as follows:

$+:V\times V\rightarrow V$ by $+(v_1,v_2)=v_1+v_2$

$\cdot~:\mathbb C\times V\rightarrow V$ by $\cdot(c_1,v_1)=c_1v_1$

The following definitions will be useful:

If $v_1+v_2=v_2+v_1$, we’ll say $+$ commutes in $V$. Similarly, if $c_1v_1=v_1c_1$, we’ll say $\cdot$ commutes in $V$.

If $(v_1+v_2)+v_3=v_1+(v_2+v_3)$, say $+$ associates in $V$. If $(c_1c_2)v_1=c_1(c_2v_1)$, say $\cdot$ associates in $V$. 

If $c_1(v_1+v_2)=c_1v_1+c_1v_2$, say $\cdot$ distributes over $+$ in $V$.

Suppose $\cdot$ commutes in $V$. Then if $(c_1+c_2)v_1=c_1v_1+c_2v_1$, say $+$ distributes over $\cdot$ in $V$.

Now, consider the following conditions on the sets $V$ and $\mathbb C$ and the functions $+$ and $\cdot~$:

$+$ and $\cdot$ commute in $V$.

$+$ and $\cdot$ associate in $V$.

There is some element named 0 in $V$ with the property that $v+0=v$ for any given $v\in V$.

Given any $v\in V$, there is some element in $V$ named $-v$ with the property that $v+(-v)=0$. 

There is an element named $1$ in $\mathbb C$ with the property that $1v=v$ for any given $v\in V$. Indeed, this is just the number $1$. 

$\cdot$ distributes over $+$ in $V$ and  $+$ distributes over $\cdot$ in $V$. 

If the sets $V$ and $\mathbb C$ together with the functions $+$ and $\cdot$ satisfy each of the six conditions above, we say that the set $V$ is a vector space over $\mathbb C$ under $+$ and $\cdot$; further, any element of a vector space $V$ will be called a vector.

[gemmacorrell]

[tastefullyoffensive]

[gemmacorrell]

[mathmajik]

Hypotrochoid Formation