A curve interpolating a converging sequence of points

Question

Suppose we have a sequence of points ${x}_n \in \mathbb{R}^n$ s.t. ${x}_n \to x_0$.

There holds the following: we can always define a curve $\phi \in C\left(\left[a,b\right], \mathbb{R}^n\cup \left[-\infty,+\infty\right]\right)$ s.t. $\forall i, \phi^{-1}\left(x_i\right) \neq \emptyset, t \in \left[a,b\right) \implies \phi\left(t\right) \neq x_0, \phi\left(b\right) = x_0$.

My proof attempt was:

1. Define a succession $\{t\}_n \subset \left[a,b\right)$ s.t. $t_1 = a, \{t\}_n \to b$
2. Define a family $\{\phi\}_n$ of curves s.t. $\phi_i\left(t_i\right)=x_i,\phi_i\left(t_{i+1}\right)=x_{i+1},\phi_i^{-1}\left(x_0\right)=\emptyset$
3. Define $\phi\left(t\right) = \left\{\begin{matrix} \phi_1\left(t\right) & t \in \left[t_1, t_2\right] \\ \vdots & \vdots \\ \phi_i\left(t\right) & t \in \left[t_i, t_{i+1}\right]\end{matrix}\right. , \phi\left(b\right)=x_0$
4. We can easily see that $\phi$ is continuos $\forall \left[c,d\right] \subset \left[a,b\right)$

The problem I was facing is in proving that $\lim_{t \to b}\phi\left(t\right)=x_0$.

---Update: Thanks @Mengchun Zhang for the answer.

Ok so I tought about a possible solution based around restricting the definition of $\phi_i$:

1. Keep the succession $\{t\}_n$ as before
2. Define the succession $\{M\}_n$, where $M_i$ is a point s.t. $||x_i - M_i|| = ||x_{i+1} - M_i||$
3. Define the succession $\{\varepsilon\}_n$, where $\varepsilon_i = \max\left(||x_i - x_0||,||x_{i+1} - x_0||\right)$
4. Define a family $\{\phi\}_n$ of curves as before with the added condition that $\phi_i: \left[t_1, t_{i+1}\right] \to B_{2\varepsilon_i}\left(M_i\right)$
5. Define $\phi$ as before

Now we have to show that $\phi$ is continuos at $b$:

1*) Suppose we have a succession $\{t^\star\}_n \in \left[a,b\right)$ converging to $b$

2*) By construction $\forall i$ $\exists j = \max\{j \in \mathbb{N}: t_j > t^\star_i\}$ and $k > j$ s.t. $t^\star_i > t_k$: so we have $t_j > t^\star_i > t_k$: by continuity of $\phi$, this means that $\phi\left(t^\star_i\right) \in \bigcup_{h = j}^k B_{2\varepsilon_h}\left(M_h\right)$

3*) Because $\{\varepsilon\}_n \to 0$ for $i,j$ big enough we have that we can define $r$ s.t. $j \leq r \leq k$ and $\varepsilon_h > \varepsilon_h \; \forall j \leq h \leq k$ which means that $\phi\left(t^\star_i\right) \in B_{2\varepsilon_r}\left(M_r\right)$

4*) For $i \to \infty$ it's easily seen that $j,k \to \infty$: by point 2*,3*) for $i$ big enough we have that $M_{r_i} + 2\varepsilon_{r_i} > \phi\left(t^\star_i\right) > M_{r_i} - 2\varepsilon_{r_i}$

5*)Because $\{M\}_n \to x_0$ and $\varepsilon_{r_i} \to 0$, for $i \to \infty$ we have $x_0 \geq \phi\left(t^\star_i\right) \geq x_0$, proving that $\{t^\star\}_n \to b \implies \{\phi\left(t^\star\right)\}_n \to x_0$

Answer 1

It might be clearer to define things explicitly. For example, let $\phi(b)=x_0$ and $\forall i\in\mathbb N^+$ define

$$\phi(t)=\lambda_ix_i+(1-\lambda_i)x_{i+1}\ \ \text{for}\ \ t\in[t_i,t_{i+1})$$ where $$\lambda_i=\frac{t_{i+1}-t}{t_{i+1}-t_i}\ \ \text{and}\ \ t_i=b+2^{1-i}(a-b).$$

Note that $\phi$ is piecewise linear with end points $\phi(t_i)=x_i$, thus $\phi$ is continuous on $[a,b)$. Now we show $\lim_{t\to b}\phi(t)=x_0$. Fix $\varepsilon>0$, since $x_n\to x_0$ there exists $N_\varepsilon\in\mathbb N^+$ such that $i> N_\varepsilon\Rightarrow|x_i-x_0|<\varepsilon$, hence there exists $\delta=2^{-N_\varepsilon}(b-a)$ such that

\begin{align*} |t-b|<\delta\ &\Rightarrow\ t>t_{N_\varepsilon}\\[.5em] &\Rightarrow\ |\phi(t)-x_0|=|\alpha_ix_i+(1-\alpha_i)x_{i+1}-x_0|\ \ \text{for some}\ \ i>N_\varepsilon\\[.5em] &\qquad\qquad\qquad\leq\alpha_i|x_i-x_0|+(1-\alpha_i)|x_{i+1}-x_0|\\[.5em] &\qquad\qquad\qquad<\alpha_i\varepsilon+(1-\alpha_i)\varepsilon\\[.5em] &\qquad\qquad\qquad=\varepsilon. \end{align*}