Proving that there is a path in set.

Question

Given that $\underline Y$ is a subspace of $\underline ℝ^2$ and defined as follows:

$$I = [0,1]$$

$$X = (\{1\} × I) ∪ \left(I × \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)\right)$$

$$Y = X \setminus \{ (0.5, 0) \}$$

How can one prove that there is no path from $(0,0)$ to $(1,0)$ using sequences?

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To show the contradiction I started by assuming that such a map exists:

$$f: \underline{I} \rightarrow \underline Y \ \text{continuous map}$$

$$f(0) = (0,0),\ \ f(1) = (1,0)$$

If $f$ is continuous, then it must hold that $f^{-1}[\ [0,0.5)\times \{0\} ] = A$ is a closed set. So $A$ contains its maximum $m$. From here on I'm not sure how to proceed. I wanted to somehow show that there is a sequence that due to $m$ being in $A$ cannot converge to $f(m) = (x,0)$.

Answer 1

You are on the right track.

So $A$ contains its maximum $m\in[0,1)$. On the other hand $f(m)=(t,0)$ for some $0\leq t<0.5$. Now take any sequence $(x_n)\subseteq [0,1]$ converging to $m$ from above, something like $x_n=m+(1-m)/n$. It follows that $f(x_n)\to f(m)$ and so if we write $f(x_n)=(a_n,b_n)$ then $a_n<0.5$ eventually, but since $m$ is the maximum of $A$ then $b_n>0$ (still convergent to $0$) whenever $a_n<0.5$.

It follows that the image of $f$ is not locally connected (around $f(m)$), by considering the open subset $\{(x,y)\in im(f)\ |\ x<1\}$, the projection onto $Y$-axis and preimages of clopen points. But any continuous function $f:[0,1]\to X$ into a Hausdorff space is a quotient map onto image. And being locally connected is preserved under quotients. This contradiction shows that no such continuous $f$ may exist.

Answer 2

Your space $X$ is a variant of the comb space. Many questions in this forum deal with it. Let us write $$X = \{1\} \times I \cup I \times N$$ with $N = \{0\} \cup \left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}$. For later use we define $N_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n \ge m \right\}$ and $N^*_m = \left\{ \frac{1}{n} \,\Bigg\vert\, n < m \right\}$.

My approach is this.

Consider a continuous $f : I \to X$. We claim that $f(I)$ intersects only finitely many of the sets $B_n = [0,\frac{3}{4}] \times \{ \frac{1}{n}\}$.
Otherwise there would exist a strictly increasing sequence $(n_k)$ of integers such that $D_k = f(I) \cap B_{n_k} \ne \emptyset$ for all $k$. Pick $t_k \in I$ such that $f(t_k) \in D_k$. The sequence $(t_k)$ has a convergent subsequence; so let us assume w.l.o.g. that $t_k \to \tau \in I$. By continuity $f(t_k) \to f(\tau)$. The second coordinate $f_2(t_k)$ of $f(t_k)$ is $\frac{1}{n_k}$, thus $f_2(t_k) \to 0$. Since the first coordinate $f_1(t_k)$ of $f(t_k)$ is always in $[0,\frac{3}{4}]$, we conclude that $f(\tau) = (\theta,0)$ for some $\theta \in [0,\frac{3}{4}]$. Obviuosly $t_k \ne \tau$ for all $k$. But $U = [0,1) \times \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)$ is an open neigborhood of $f(\tau) = (\theta,0)$ in $X$, thus there exists $\epsilon > 0$ such that $f(t) \in U$ for $\lvert t - \tau \rvert < \epsilon$. Choose $k$ such that $\lvert t_k - \tau \rvert < \epsilon$. Hence we get a path in $U$ connecting $f(\tau)$ and $f(t_k)$. This a contradiction because these points lie in different path components of $U$.

Now assume there is a path in $Y$ from $(0,0)$ to $(1,0)$. Then there also is a path $f$ from $(0,0)$ to $(1,1)$ because $(1,0)$ and $(1,1)$ lie in the same path component of $Y$. We know that $f(I) \subset Y_m = \{1\} \times I \cup I \times N^*_m \cup [\frac{3}{4},1] \times N_m \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$ for sufficiently large $m$. But $Y' = \{1\} \times I \cup [\frac{3}{4},1] \times N \cup (I \setminus \{\frac{1}{2} \}) \times \{0\}$ is a retract of $Y$, hence $f$ induces a path $f'$ in $Y'$ from $(0,0)$ to $(1,1)$. The coordinate function $f_1'$ is a continuos real-valued function such that $f_1'(0) = 0, \ f_1'(1) = 1$. The IVT says that $f_1'(t) = \frac{1}{2}$ for some $t$ between $0$ and $1$. This is a contradiction because $Y'$ does not contain any point whose first coordinate is $\frac{1}{2}$.