# Counterexamples in Topology and Algebra-I

This is a part of a series of posts which will deal with different types of counterexamples in topology and algebra.

So, what exactly is a counterexample in mathematics. It must be an example which disproves a conjecture but here I view it more broadly. A counterexample is simply an interesting example which deviates from the common results associated with various ‘common’ spaces and objects(eg: Euclidean space and real line) which most students encounter.

The mathematical intuition, so to speak, is a truly fascinating aspect of studying mathematics. While the ability to condense a block full of mathematical jargon and symbols into a single precise ‘picture’  or performing the opposite, that is, delineating mysterious mathematical intuition into pages of rigorous proof and exposition may sound quite romantic, there’s quite often the sporadic chance of failing to translate intuition to reality-the counterexample.

Topology is a notorious example of this unfortunate possibility, especially in instances of generalizing finite-dimensional results. But the mathematicians must persevere, transcending spatial imagery while still grounding his work, in some manner or another, on the basic axioms of mathematics and the Euclidean space.

Starting from the well-known Hausdorff condition and the distinction between the box and product topology, I often question whether my incessant quest for abstraction is leading me astray either to the swamp of unscrupulous simplistic intuition or to a bleak cave of incoherent definitions and theory where I forever sit, merely content with knowing but not understanding-the daring step to rebuild my intuition, a lost dream.

Terence Tao has a rather interesting take on the topic. In his blog, he discussed how mathematicians learn and evolve from their naive mathematical thinking but finally revert back to it, albeit with new insight and wisdom which they previously lacked- a refined intuition of sorts.

Counterexamples are a great way to properly ‘train’ our intuition and unearth our mathematical insecurities.

Enough rambling, here are some pretty cool counter-examples to think of.

## THE DELETED COMB SPACE

A topological space $X$ is connected if it is not possible to write it as the union of two open disjoint subsets of $X$.

A topological space $X$ is path-connected if for every pair of points $x,y$ in X, there exists a continuous function/path $p:[0,1] \mapsto X$ such that $p(0)=x,p(1)=y$.

Now, it is easy to see that a path-connected space is always connected but the converse need not be true.

The best example of the failure of the converse is the Topologist’s sine curve defined as $X=\{sin(\frac{1}{x})|x>0\} \cup (0,0)$.

The Deleted Comb Space is another amazing example of this failure.

It is defined as $D=\{[0,1] \times \{0\} \} \cup \bigcup\limits_{n=1}^{\infty} (\{\frac{1}{n}\} \times [0,1]) \cup (0,1)$

Proving that it is connected is pretty simple. Now, to prove that $D$ is not path-connected, we’ll show that no path connects  $(0,1)$ to any other point on $D$

First, we assume that there exists a path $p:[0,1] \mapsto D$ such that $p(0)=(0,1)$. Then, we basically show that $p(t)=(0,1) \forall t \in [0,1]$ which basically means that the function can’t ‘move past’ $(0,1)$.

Consider $A=\{t \in [0,1]| f(t)=(0,1) \}$. The proof is complete when we show that A is both closed and open(since the only clopen sets of $\mathbb{R}$ are $\mathbb{R}$ and the void set).

By definition, $A$ is closed since $A=p^{-1}(0,1)$ .

Choose some $t_{0} \in A$. By the continuity of $p$, there exists some $\delta >0$ such that $B(t_{0};\delta) \subset f^{-1}(V)$ where $V$ is an open set containing $(0,1)$ in $A$ under the subspace topology such that $V$ doesn’t intersect the X-axis.

Now, the projection function $\pi_{1}:\mathbb{R}^{2} \mapsto \mathbb{R}$ defined by $\pi_{1}(x,y)=x$ is continuous. Since the composition of two continuous functions is also continuous, the function $f:B(t_{0};\delta) \mapsto \mathbb{R}$ given by $f(t)=\pi_{1}(p(t))$ is continuous.

Let $I=B(t_{0};\delta) \cap [0,1]$. $I$ is clearly connected.So, $f(I)$ is also connected.

This means that $f(I) \subset X$ where $X=\bigcup\limits_{n=1}^{\infty} \{\frac{1}{n}\}$. Any connected set $C$of $X$ contains some point of the form $\frac{1}{n}$. But $\frac{1}{n}$ is both closed and open in $X$. So, $C=\frac{1}{n} \cup (C-\frac{1}{n})$ is a separation of $C$. Since $C$ is connected,$C-\frac{1}{n}=\phi$ hence $C=\frac{1}{n}$ for some $n \in Z^{+}$.

Therefore,$f(I)$ is a one point set. Since $t_{0} \in I, f(I)=f(t_{0})=\pi_{1}(p(t_{0})=\pi_{1}(0,1)=0$. $f(I)=\{0\}$. So, $I \subset A$.

This means that $A$ is open as for every $t_{0} \in A$, there exists an neigbourhood $I$ of $t_{0}$ in $A$. The only clopen non-empty subset of $[0,1]$ is $[0,1]$ itself. So,$A=[0,1]$ and the function $p$ is a constant function which is a contradiction.