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  1. Show $f:\mathbb{R}^2\to\mathbb{R}, f(x_1,x_2)=x_1$ is continuous.

How should I proceed?

Fix $\epsilon>0 |f(x_1,x_2)-f(x_3,x_4)|<\epsilon\Rightarrow |x_1-x_3|<\epsilon\Rightarrow$ choose $\delta=\epsilon$

  1. $A$ is open in $\mathbb{R}$, $B=\{(x_1,x_2): x_1\in A\}$, show $B$ is open in $\mathbb{R}^2$.

If (1) is shown then I can directly tell inverse image of open set is open. thanks for helping

Myshkin
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2 Answers2

1

You are mostly right, just need to sort notation out.

Continuity happens at a point, and if it happens at every point then it happens in the space. So first we need to fix a point, say $z = (z_1,z_2) \in \mathbb R^2$.

Now, the definition of continuity is : For all $\epsilon > 0$, there is $\delta > 0$ such that $||\mathbf{x} - z|| \leq \delta \implies |f(\mathbf x)-f(z)| < \epsilon$. (Here, $\mathbf x \in \mathbb R^2$). $\mathbb R^2$ is normally equipped with Euclidean norm (although you can show that all norms are equivalent, so we can switch norms), so I will persist with it here.

Now, if $\mathbf x= (x_1,x_2)$, then $|f(\mathbf x)-f(z)| = |x_1 - z_1|$. On the other hand, $||\mathbf x - z|| = \sqrt{(x_1 - z_1)^2 + (x_2 - z_2)^2}$.

You can see that $$||\mathbf x - z || < \epsilon \implies |x_1 - z_1| < \epsilon$$

(Square both sides, use the fact that squares are non-negative, and then take square root again)

Hence, it follows that $f$ is continuous at $z$. Now, since $z$ was arbitrary, we get that $f$ is continuous on $\mathbb R^2$.

From this, the second part follows easily, since $B = f^{-1} (A)$, where $A$ is open in $\mathbb R$, and $f$ is continuous (on $\mathbb R^2$).

Sarvesh Ravichandran Iyer
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The first implication looks suspicious in your work, which can be somewhat expanded to the degree that your reader can tell you really know your stuff.

Let $(c_{1},c_{2}) \in \mathbb{R}^{2}$; let $\varepsilon > 0$. Note that $|f(x_{1},x_{2}) - f(c_{1},c_{2})| = |x_{1} - c_{1}|$ for all $(x_{1},x_{2}) \in \mathbb{R}^{2}$. If in addition $(x_{1},x_{2})$ is within the open ball of center $(c_{1},c_{2})$ and radius $\varepsilon$, then $\sqrt{(x_{1}-c_{1})^{2} + (x_{2}-c_{2})^{2}} < \varepsilon$, implying that $(x_{1}-c_{1})^{2} < \varepsilon^{2}$, implying that $|x_{1}-c_{1}| < \varepsilon$. So taking $\delta := \varepsilon$ suffices.

Yes
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