If $ y \in [0,1] \times [0,1]$, is $[0,1] \times [0,1]-y$ connected?

Question

$(a)$ If $ y \in [0,1] \times [0,1]$, then prove/disprove that $[0,1] \times [0,1]-\{y\}$ is connected.

$(b)$ Hence, prove that a continuous, onto mapping $f :[0,1] \rightarrow [0,1] \times [0,1]$ can never be one-one

Attempt:

For part $(a) :$

I have a graphical situation in my mind. $[0,1] \times [0,1]$ represents a square in $\mathbb R^2$. Hence, even if we remove a single point from this square, we get an arc connected structure.

But, how do we write a formal proof?

For part $(b)$ : Suppose $f$ is one-one and onto. Then $f^{-1}$ exists and is continuous.

Unfortunately, I don't have much idea on how to move forward. Please suggest me an approach.

Thank you for your help..

Answer 1

a) You nearly have the proof. Given two points in $X=[0,1]\times [0,1]-\{y\}$, say $p_1=(a,b)$ and $p_2=(c,d)$, how would you connect them with an arc $\gamma:[0,1]\to X$? Well, if the removed point $y$ does not lie on the line connecting the two points, then $\gamma (t)=tp_1+(1-t)p_2$ works. Otherwise, you'll need to adapt the path a bit.

b) If a continuous and bijective function $f:[0,1]\to [0,1]\times [0,1]$ exists, then its restriction to $[0,1]-\{1/2\}$ would give you a continuous bijective function between the non-connected space $[0,1]-\{1/2\}$ and the connected space $X$ (for which $y$?). Why is that impossible?

Answer 2

Note: it is easy to connect any two points in this set with a continuous path. This shows that the space is connected.

Answer 3

Here is how you should think about connectivity in this setting:

Definition: A topological space $X$ is called path connected (or arc connected) if, given any $a, b \in X$, there exists a continuous function $f:[0,1] \rightarrow X$ such that $f(0) = a$, $f(1) = b$.

Lemma: $[0,1]$ is connected.

Proposition: If $X$ is path connected, then $X$ is connected.

Sketch of proof: Assume $X$ is path connected but that there exists a separation of $X$, i.e. there are nonempty open sets $U$, $V$ in $X$ such that $U \cap V = \varnothing$, $U \cup V = X$. Pick $a \in U$, $b \in V$, and find a path $f$ from $a$ to $b$. Consider $f^{-1}(U)$ and $f^{-1}(V)$ in $[0,1]$ and derive a contradiction of the lemma.

A warning: there do exist spaces that are connected but not path connected. These tend to be rather pathological, though.